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MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry

Environmental Chemistry Important Questions

Environmental Chemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Source of non – pollution energy:
(a) Fossil fuel
(b) Sun
(c) Gasoline
(d) Nuclear energy
Answer:
(b) Sun

Question 2.
Which radiations manufacture O₃:
(a) Ultra – violet
(b) Visible
(c) Infra – red
(d) Radio waves
Answer:
(a) Ultra – violet

Question 3.
Which radiations provide Green house effect:
(a) Infra – red
(b) Visible
(c) Ultra – violet
(d) X – rays
Answer:
(c) Ultra – violet

Question 4.
PAN is responsible for:
(a) Depletion of ozone layer
(b) For smog
(c) For Acid rain
(d) Poisonous food
Answer:
(b) For smog

Question 5.
Which is not an air pollutant:
(a) H₂
(b) H₂S
(c) NO₂
(d) O₃
Answer:
(a) H₂

Question 6.
Air pollutant released from jet aeroplanes in the form of an air pollutant:
(a) Photochemical oxidant
(b) Photochemical reductant
(c) Aerosol
(d) Physical pollutant
Answer:
(c) Aerosol

Question 7.
Is not present in Acid rain:
(a) H₂SO₄
(b) HNO₃
(c) H₂SO₃
(d) CH₃COOH
Answer:
(d) CH₃COOH

Question 8.
Is responsible for disease of lungs:
(a) O₂
(b) N₂
(c) CO₂
(d) SO₂
Answer:
(d) SO₂

Question 9.
O₃ is manufactured in:
(a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Thermosphere
Answer:
(b) Stratosphere

Question 10.
For acid rain ‘sink’ is:
(a) Leaves
(b) Reservoir
(c) Lime stone
(d) CO₂
Answer:
(c) Lime stone

Question 11.
Primary pollutant is:
(a) SO₃
(b) NO₂
(c) N₂O
(d) NO
Answer:
(d) NO

Question 12.
Most dangerous is:
(a) Smoke
(b) Dust
(c) Smog
(d) NO
Answer:
(c) Smog

Question 2.
Fill in the blanks:

1. The air pollutant released by jet aeroplanes in the form of fluro carbon is …………………………..
2. D.D.T. is …………………….. poisonous pollutant as compared to B.H.C.
3. O₃ is formed in the ……………………… level of atmosphere.
4. A definite tolerable level of pollutants in the environment is expressed by …………………………….
5. Maximum air pollutants are present in …………………………….. level of the atmosphere.
6. Ozone layer prevent us from …………………………….. rays.
7. Oxides of …………………………. and …………………………….. cause acid rain.
8. …………………………… is the main cause of ozone layer depletion.
9. ………………………….. gas is responsible for Green house effect.
10. Substance produces pollution is known as ……………………………..
11. ……………………………… is responsible for lung diseases.
12. SO₂ pollutant is responsible for the disease of ……………………………..

Answer:

1. Aerosol
2. More
3. Stratosphere
4. T.L.V.
5. Troposphere
6. Ultra – violet
7. Nitrogen, sulphur
8. C.F.C
9. CO₂
10. Pollutant
11. Photochemical smog
12. Asthma

Question 3.
Answer in one word/sentence:

1. Name the person who started Chipko Aandolan for the conservation of forest?
2. Region of the atmosphere where living beings exist is known as?
3. What is the name of compound responsible for hole in ozone layer?
4. What is smoke containing fog known as?
5. What is rain water containing small amount of sulphuric acid and nitric acid known as?
6. Which chemistry produces environment friendly chemicals having minimum contribution in pollution?
7. What is decrease in density of ozone gas due to chlorofluorocarbon compound in atmosphere known as?
8. What is the maximum quantity of pollutant having no effect on receptor known as?
9. When do we celebrate World Environment Day?
10. What is PAN?
11. Name the largest sink of earth?
12. When does Bhopal Gas Tragedy occured?
13. Name the disease due to water pollution?

Answer:

1. Sunderlal Bahuguna
2. Troposphere
3. Chlorofluorocarbon
4. Smog
5. Acid rain
6. Green chemistry
7. Ozone hole
8. Threshold limit value
9. 5th June
10. Peroxyacetyl nitrite/ photochemical smog
11. Ocean
12. Midnight of 2nd and 3rd December 1984
13. Jaundice, Diarrhoea

Question 4.
Match the following:
[I]

Answer:

1. (e)
2. (c)
3. (b)
4. (d)
5. (a)

[II]

Answer:

1. (c)
2. (d)
3. (a)
4. (b)
5. (e)

[III]

Answer:

1. (b)
2. (a)
3. (d)
4. (c)

Environmental Chemistry Very Short Answer Type Questions

Question 1.
Write the definition of pollutants?
Answer:
Such substances whose amount in the environment increases more than required and cause harmful effect on human, animal and plant kingdom are called pollutants. Like: CO, CO₂, NO, NO₂, SO₂ etc.

Question 2.
Write names of two air pollutants?
Answer:
SO₂, SO₃.

Question 3.
Name two pollutants which depletes ozone layer?
Answer:

1. Cycle of nitric oxide (NO) and
2. C.F.C. (Chlorofluoro carbon), in which C.F.C. is main.

Question 4.
Ozone is found in?
Answer:
Stratosphere.

Question 5.
Write the two health problems caused by SO₂?
Answer:

1. SO₂ affects the respiratory canal and lungs due to which various health diseases are caused like cancer.
2. Acid rain caused due to SO₂ produces boils on the skin.

Question 6.
What is Acid rain?
Answer:
Various gaseous pollutant present in the atmosphere like SO₂, SO₃, NO₂, NO dissolve in rain drop.
These drops fall with rain and are called acid rain.
SO₂ + H₂O → ₂HSO₃
SO₃ + H₂O → H₂SO₄.

Question 7.
Name any two green house gases?
Answer:
C.F.C. and CO₂.

Question 8.
What is P.A.N.?
Answer:
It is peroxy acyl nitrate which is a photochemical smog.

Question 9.
What is C.F.C.?
Answer:
It is chlorofluoro carbon which is the main cause of depletion of ozone layer.

Question 10.
What is green chemistry?
Answer:
A technique to check pollution in which such chemical reactions are suggested which do not cause pollution and if pollution spreads then it can be destroyed. This is called green chemistry.

Question 11.
What is TLV?
Answer:
A definite value of pollutants can be tolerated. This is expressed by TLV. TLV means ‘Threshold Limit Value’.

Question 12.
What are particulate pollutants?
Answer:
The pollutants mixed up with air in liquid or solid state in such a manner that the remain suspended for a long time is called particulates.

Question 13.
What is specimasion?
Answer:
Many pollutants can be made from element. The method of determining which of the product is more dangerous is called specimasion.

Question 14.
What is sink?
Answer:
Sink is that in which the substance totally gets consumed and then also there is no effect on sink.

Question 15.
Name the biggest sink of the earth?
Answer:
The biggest sink of the earth is the sea.

Question 16.
What is green house effect?
Answer:
The heating of atmosphere due to absorption of infrared radiation by carbon dioxide and other gases is called green house effect.

Question 17.
Explain the mechanism of acid rain?
Answer:
All pollutant gases spread as pollutant particles in the form of smoke by the burning of fossil fuels and other fuels. Due to high temperature of industries and other engines, oxides of nitrogen spread in the atmosphere by the combination of N₂ and O₂. These gases mix in rain drops form acid and then fall on the earth as acid rain drops. This is called acid rain.

Question 18.
The gas which gets leak in 1984 in Bhopal is?
Answer:
CH₃ – N = C = O Methyl isocyanate.

Question 19.
What is the main component for the depletion of ozone layer?
Answer:
Chlorofluorocarbon.

Question 20.
Name the person who started “Chipko Andolan” for forest conservation?
Answer:
Sunderlal Bahuguna.

Question 21.
Write the name of disease caused by water pollution?
Answer:
Cholera, Typhoid, Joindiss etc.

Question 22.
Name the medicinal plant which is helpful in controlling pollution and useful in skin diseases?
Answer:
Neem (Azaderecta indica).

Question 23.
Which gas is responsible for green house effect?
Answer:
Carbon dioxide (CO₂).

Question 24.
What is the main sink of CO pollutant?
Answer:
Biological molecules present in soil.

Question 25.
Write the name of four methods useful in Green chemistry?
Answer:

1. Use of sunlight
2. Micro oven
3. Micro waves
4. Sound waves
5. Use of enzyme.

Question 26.
Which pollutant is responsible for smog?
Answer:
PAN (Peroxy acyl nitrate).

Question 27.
What is sink for acid rain?
Answer:
Lime stone (Marble).

Question 28.
Which sphere is present near to earth?
Answer:
Troposphere.

Question 29.
Full form of B.H.C. is?
Answer:
Benzene Hexa chloride.

Question 30.
Which light is responsible for skin cancer?
Answer:
Ultra – violet light (UV – light).

Question 31.
What is responsible for lung diseases?
Answer:
Sulphur dioxide (SO₂).

Question 32.
What is the main source of emmision of CO?
Answer:
Vehicles.

Question 33.
Oxides of which elements are responsible for acid rain?
Answer:
Oxides of nitrogen and sulphur.

Question 34.
Maximum pollution occurs in which sphere?
Answers:
Troposphere.

Question 35.
What is the main constituent for ozone depletion?
Answer:
C.F.C. (Chlorofluoro carbon).

Question 36.
Which radiation give green house effect?
Answer:
Infrared radiations (IR).

Question 37.
The formation of ozone takes place where in the atmosphere?
Answer:
Stratosphere.

Environmental Chemistry Short Answer Type Questions – I

Question 1.
Define Green Chemistry?
Answer:
Green chemistry is the branch of science in which study of effects of chemicals (like : Origin, transportation, reactions etc.) on environment is studied.

Question 2.
Explain Tropospheric pollution?
Answer:
Tropospheric pollution occurs due to unwanted solids and gas particles present in the air. The pollution occurs due to following two substances:

1. Gaseous air pollutants:
They are sulphur, nitrogen and CO₂, H₂S, hydrocarbon, ozone and other oxidising agents.

2. Particulates:
They are dust, fog, smoke etc.

Question 3.
Which gases are responsible for green house effect?
Answer:
Main gases are CO₂, methane, water vapour, nitrous oxide, chlorofluoro carbon (CFC) and ozone.

Question 4.
What do you mean by BOD (Biochemical oxygen demand)?
Answer:
The total amount of oxygen consumed by microorganism in decomposing the wastes present in a certain volume of a sample of water.

Question 5.
Due to green house effect temperature of the earth is increasing? Which substances are responsible for it?
Answer:
Green house gases like CO, methane, nitrous oxide, ozone and chlorofluoro carbons are responsible for green house effect.

Question 6.
Ozone is a toxic gas and is a strong oxidizing agent even then its presence in the stratosphere is very important Explain what would happen if ozone from this region completely removed?
Answer:
The ozone layer acts as a protective umbrella and does not allow the harmful UV radiations to reach the earth’s surface.(MPBoardSolutions.com) If ozone is completely removed from the stratosphere, the UV radiations will fall directly on the humans, causing skin cancer and on the plants affecting plant proteins.

Question 7.
What are the sources of dissolved oxygen in water?
Answer:

1. Photosynthesis
2. Natural aeration
3. Artificial aeration.

Question 8.
Dissolved Oxygen in water is very important for aquatic life. What processes are responsible for the reduction of dissolved oxygen in water?
Answer:
Dissolve oxygen is essential for sustaining animal and plant life in any aquatic system. The wastes such as domestic, industrial and biodegradable organic compounds are oxygen demanding wastes. These are decomposed by the bacterial population which in turn decreases the oxygen from water.

Question 9.
What are Biodegradable and non – biodegradable pollutants?
Answer:

• Biodegradable pollutants: They can be degrade by microorganisms.
• Example: Sewage, dungs of animals, fruits and vegetable peels etc.
• Non – biodegradable pollutants: They cannot degrade by microorganisms.
• Example: Mercury, Lead, DDT, glass, plastic etc.

Question 10.
What is pollution?
Answer:
Environmental pollution is the effect of undesirable changes in our surroundings that have harmful effects on plants, animals and human beings.

Question 11.
What is pollutant?
Answer:
A substance present in the environment in greater proportion than its natural abundance and resulting into harmful effects, is called a pollutant.

Question 12.
What are contaminants?
Answer:
Some substances which are not present in the environment, but are released in the environment as a result of chemical activities lead to pollution. Such substances are called contaminants. Example: Methyl isocyanate gas (CH₃NCO).

Question 13.
Write the chemical name of the gases depleting ozone?
Answer:
Nitric oxide, atmospheric oxygen and chlorofluoro carbon are responsible.

Question 14.
Which are green house gases?
Answer:
CO₂, ozone and water vapours are green house gases. They have the tendency to absorb IR radiations.

Question 15.
What is polluted air?
Answer:
If some underisable substances get added in the air which affects the health of the organisms, then such air is called polluted air.

Question 16.
Why there is ozone depletion over Antarctica?
Answer:
In the stratosphere compounds formed are converted back into chlorine free radical which deplete ozone layer.

Question 17.
What is the importance of BOD measurement of any water sample?
Answer:
BOD is the measurement of pollution caused by organic biodegradable substances in water sample. The less value of BOD tells that less amount of organic effluent is present in water.

Question 18.
Oxidation of sulphur dioxide into sulphur trioxide in the absence of a catalyst is a slow process but this oxidation occurs easily in the atmosphere. Explain how does this happen. Give chemical reaction for the conversion of SO₂ into SO₃
Answer:
The oxidation of sulphur dioxide into sulphur trioxide can occur both photochemically or non – photochemically. In the near ultraviolet region, the SO₂ molecules react with ozone photochemically.
SO₂ + O₃ \(\underrightarrow { h\nu } \) SO₃ + O₂
2SO₂ + O₂ \(\underrightarrow { h\nu } \) 2SO₃
Non – photochemically, SO₂ may be oxidised by molecular oxygen in presence of dust and soot particles.
2SO₂ + O₂ \(\underrightarrow { Particulates } \) 2SO₃

Question 19.
How is ozone formed in stratosphere?
Answer:
Ozone in the stratosphere is a product of (UV) radiations acting on dioxygen (O₂) molecules. The (UV) radiations split apart molecular oxygen into free oxygen (O) atoms. These oxygen atoms combine with the molecular oxygen to form ozone.

Question 20.
What is chlorosis?
Answer:
Chlorophyll in plants is formed slowly. This is due to presence of SO. This pollution is called chlorosis.

Question 21.
What is Metathesis?
Answer:
Metathesis is the name of that science, in which the study of application of chemical methods for general persons is studied.

Question 22.
What are primary and secondary air pollutants?
Answer:
Primary air pollutants are those which remain as such after their formulation e.g. NO, while secondary pollutants are formed as a result of chemical interaction between primary air pollutants. Example: PAN.

Question 23.
What is photochemical smog?
Answer:
Photochemical smog:
It is formed by photochemical reactions involving solar radiations. The principal constituents are O₃, NO₂ and some photochemical oxidants. It is also called Los Angeles smog.

Question 24.
What is acid rain?
Answer:
Acid rain:
It is the rain water containing sulphuric acid, nitric acid and small amount of hydrochloric acid which are formed from the oxides of sulphur and nitrogen present in the air as pollutants and has a pH of 4 – 5.
CO₂ + H₂O → H₂CO₃
SO₃ + H₂O → H₂SO₄

Question 25.
When CO₂ is called harmful gas?
Answer:
Normal amount of CO₂ in atmosphere is not harmful. Whereas organisms and plants prepare their food with the help of it. But, when the amount of CO₂ increases due to various process then it alters the environmental balance and become harmful.

Question 26.
What is the role of CO₂ in the “Green house effect”?
Answer:
Water vapours are present only near the earth atmosphere but ozone found very far from the earth and CO₂ is found everywhere in the atmosphere. (MPBoardSolutions.com) So for green house effect, CO₂ is more responsible because CO₂ has the tendency to absorb IR radiations. Due to this green house effect produced.

Question 27.
Why acid rain is harmful for Tajmahai?
Answer:
Tajmahai is made up of marbles (CaCO₃). Acid rain contains H₂SO₄ in dilute state. It reacts with marble of Tajmahai and make it discoloured and lustreless.

Question 28.
What is Fly ash pollution?
Answer:
The smaller ash particles are formed when fossil fuel is burnt. Gases produced during burning take the ash particles and pollute the atmosphere. The pollution is called fly ash pollution.

Question 29.
What is contaminated water? Name of diseases occurs due to it?
Answer:
The water which contains dissolved organic matters and salts and micro – organisms is called contaminated water.
Diseases are:

1. Diarrhoea
2. Typhoid
3. Skin diseases etc.

Question 30.
What is Global warming?
Answer:
This increase in average temperature of global air due to increased green house effect is called ‘global warming’.

Question 31.
What is Ionosphere?
Answer:
It is also 40 km thick layer at an altitude of 50 km from the earth surface. It is also called mesosphere. Various ionic reactions taking place in ionosphere is:
O₂ + \(\overset { \bullet }{ O } \) → O₂⁺ + O
O⁺ + N₂ → NO⁺ + N
N₂⁺ + O₂ → N₂ + O₂⁺

Environmental Chemistry Short Answer Type Questions – II

Question 1.
How is the poisonous effect of CO produced on man and animals?
Answer:
CO has poisonous effect on man and plants.
1. It combines with haemoglobin of the blood more strongly than oxygen.
CO + Hb → CO – Hb (Carboxy haemoglobin).

As a result of this amount of haemoglobin available in the blood for the transport of oxygen to the body cells decreases. The normal metabolism is thus, impaired due to less O₂ level. This will cause suffocation and will ultimately lead to death. (MPBoardSolutions.com) Carbon monoxide if present in air can cause mental impairment, respiratory problems, muscular weakness and dizziness.

2. A high concentration of CO (100 ppm or more) will harmfully affect the plants causing leaf drop; reduction of leaf size and premature aging etc.

Question 2.
Write the harmful effects of SO₂ (Sulphur dioxide)?
Answer:
Harmful effects of SO₂ are:

1. SO, affects respiratory tract producing nose, eye and lung irritation. It has been reported that lower concentration of SO₂ causes respiratory weakness.
If present at a concentration of only 2.5 ppm in the environment, then also it leads to dangerous diseases like bronchitis and lung cancer etc.

2. SO₃ produce harmful effect on buildings made of marble and lime stones (CaCO₃). The gas released from Mathura oil refinery is harmful for the Tajmahal.

3. High concentration of SO₂, leads to stiffness of flower buds which eventually fall off from plants.

4. Air polluted by oxides of sulphur enhances the corrosion of metals like copper, zinc, iron etc.

Question 3.
Write the harmful effects of nitrogen dioxide?
Answer:
Nitrogen dioxide is harmful and poisonous. It produces the following harmful effects:

1. It reacts with the ozone present in the atmosphere and decreases its density.
2. Oxides of nitrogen are responsible for the production of photochemical smog.
3. Oxides of nitrogen cause harmful effects on textile fibres like nylon, rayon, cotton fibres etc. NO2 causes cracks in rubber.
4. Increase in concentration of NO₂ in the atmosphere is harmful for plants. It leads to leaf spotting, retards photosynthetic activity, retards plant growth etc.
5. NO₂ creates problems in human respiration and leads to bronchitis.

Question 4.
A farmer was using pesticides on his farm. He used the product of his farm as food for rearing fishes. He was told that fishes were not fit for human consumption because large amount of pesticides had accumulated in the tissues of Fishes. Explain, how did this happen?
Answer:
Pesticides are organic compounds which are used to protect plants from pests. These are mild poisons. These pesticides stick to the plants and also flow into lakes along with the rain water. (MPBoardSolutions.com) Rearing fishes when consume these plants as their food the poisonous pesticides accumulate in the tissues of fishes. Thus, these fishes are not fit for human consumption.

Question 5.
For dry cleaning, in the place of tetrachloroethene, liquefied carbon dioxide with suitable detergent is an alternative solvent. What type of harm to the environment will be prevented by stopping use of tetrachloroethane? Will use of liquefied carbon dioxide with detergent be completely safe from the point of view of pollution? Explain?
Answer:
1. Tetrachloroethene (Cl₂C = CCl₂) is suspected to be carcinogenic and contaminates the ground water. This harmful effect will be prevented by using liquefied CO₂ along with suitable detergent.

2. Use of liquefied CO₂ along with detergent will not be completely safe because detergents also cause pollution as most of the detergents are non – biodegradable. Also, liquefied CO₂ will ultimately enter into the atmosphere and contribute to the green house effect.

Question 6.
What are the harmful effect of Green house effect?
Answer:
Though green house effect was beneficial in maintaining a livable temperature on earth. But excessive CO₂ in the atmosphere due to deforestation and large scale burning of fossil fuel has disturbed the natural balance in favour of higher green house effect. This has led to an increase in average temperature of the earth from 0.3 to 0.6°C over the past century. This increase in average temperature of global air due to increased green house effect is called ‘Global warming’.

The atmospheric CO₂ level is expected to become double sometimes between 2050 – 2150 with a corresponding increase in global temperature from 1 to 3°C. Besides CO₂, other green house gases are methane, water vapour, nitrous oxides, CFCs (Chlorofluorocarbons) and ozone. Methane is produced naturally when vegetation is burnt, digested or rotted in the absence of oxygen. (MPBoardSolutions.com) Large amount of methane are released in paddy fields, coal mines, from rotting garbage dumps and by fossil fuels.

CFCs are man made industrial chemicals used in air conditioning etc. CFCs are also damaging the ozone layer. Nitrous oxide occurs naturally in the environment. In recent years, their quantities have increased significantly due to use of chemicals, fertilizers and the burning of fossil fuels.

Harmful Effects of Global Warming:

• There will be rise in sea level due to increased rate of melting of glaciers. Sea level may rise by 0.5 to 1.5 m during the next 50 to 100 years if present rise in CO₂ continues. This will result in flood and loss of soil particularly in coastal areas.
• Higher global temperature is likely to effect the whole ecosystem by disturbing the life cycle of certain micro and macro organisms.
• Higher temperature is likely to increase incidence of infectious diseases such as malaria, dengue, yellow fever and sleeping sickness.

Question 7.
Write the reasons of water pollution?
Answer:
The sources of water pollution are as follows:

1. Organic pollutants like manure wastes from food processing, rags, paper discards etc.
2. Industrial wastes.
3. Detergents and Fertilizers: The detergents are best available mode for the growth of bacteria.
4. Pollution of water takes place through acids.

Question 8.
How is artificial green house prepared?
Answer:
Synthetic green house:
In nature, coating of CO₂ is forming green house, but synthetic green house can be synthesized by studying its mechanism.

Actually the transparent glass roof and wall of the glass house allow sun rays to pass through and strike the surface of the house. The reflected radiation is of longer wavelength than the incident radiation. A significant portion of reflected radiation absorbed by glass. (MPBoardSolutions.com) As radiation of longer wavelength (Infrared radiation) generates heat, this causes rise in temperature inside the glass house. An effect similar to one in glass house is responsible for keeping the earth’s surface warmer.

Question 9.
For your agricultural field or garden you have developed a compost producing pit Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce?
Answer:
The compost is very useful for agriculture as a fertilizer. But the compost pro-ducing pits may give bad odour and flies. Therefore, the compost producing pit should be set up at a suitable place or in a bin to protect ourselves from bad odour and flies. It must be kept covered so that flies cannot enter into it and there is not much a bad odour.

Question 10.
How house effluents can be used as manure?
Answer:
Waste Management of Household waste:
All the solid household waste should be put in the household garbage box/bin. This should be then put into the community bins so that the municipal workers can take it in their vehicles to the disposable site. Here, the garbage is separated into biodegradable and non – biodegradable materials. (MPBoardSolutions.com) The biodegradable waste is deposited in the land fills. With the passage of time, it is converted into manure compost. Remember that if the waste is not collected in the garbage bins, it may find its way to sewers and some may to eaten up by the cattle.

The non – biodegradable waste like polythene bags, metal scrap, etc. choke the sewers. The polythene bags, if swallowed by cattle, can result into their death. The best way to manage domestic waste is to keep two garbage bins, one for the biodegradable (Non – recyclable) and the other for non – biodegradable (Recyclable) which can be sold to the vendor/dealer.

Question 11.
What do you mean by green chemistry? How will it help in decrease environmental pollutions?
Answer:
By green chemistry we mean a strategy to design chemical process which neither use toxic chemicals nor release the same to the atmosphere. It also means to develop methods of using raw materials more efficiently and generating less wastes.

The creative and innovative skills of green chemistry has developed many new environmental friendly processes, analytical tools, reaction conditions and catalysts etc. A few of these achievement may be listed as follows:

1. Development of new method to improve the yield of ibuprofen upto 99%.
2. Chlorofluorocarbon used as blowing agents for polystyrene foam (Thermocol) sheets have been replaced by CO2.
3. A new technique of catalytic dehydrogenation of ‘diethanolamine’ produces an environment friendly herbicide. This process has avoided the use of highly toxic cyanide and formaldehyde.
4. Organotin: A common antifouling compound used by sea marines has been replaced by a rapidly degradable compound called ‘sea nine’.

Question 12.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon monoxide is highly poisonous in nature. It combines readily with haemoglobin (It has more affinity than oxygen). Due to the formation of carboxyhaemoglobin, the quantity of oxygen to the body cell get reduced i.e. (MPBoardSolutions.com) CO reduces the oxygen carrying capacity of the blood and this leads to oxygen starvation (Anoxia). The deficiency of oxygen produces headache, dizziness, choking cardiac and pulmonary complications leading to paralysis and death. CO₂ does not combine with haemoglobin. However, it is a green house gas and helps in global warming. Hence, it is less dangerous pollutant.

Question 13.
What would have happened if the green house gases were totally missing in the earth’s atmosphere? Discuss?
Answer:
The solar energy radiate back from earth surface is absorbed by the green house gases (i.e. CO₂, CH₄, O₃, CFC and water vapour) present near the earth surface. Thus, they heat up the atmosphere near the earth’? surface and keep it warm. (MPBoardSolutions.com) As a result, they keep the temperature of the earth constant and help in the growth of plants and existence of life on the earth. If there were no green house gases, there would have no vegetation and life on the earth.

Question 14.
Statues and monuments in India are affected by acid rain. How?
Answer:
Statues and monuments are generally made of marble (Taj Mahal). The acid rain contains H₂SO₄ which attacks the marble.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
As a result, the monuments (Taj mahal) are being slowly corroded and the marble is getting discoloured and lustreless.

Question 15.
What are the harmful effect of water pollution? How they can be controlled?
Answer:
The harmful effects of water pollution are:

1. Due to intake of polluted water many diseases like typhoid, dysentry etc. occur.
2. Due to effluents the amount of dissolved oxygen in water decreases.
3. Due to presence of soap and detergent effluents the water become poisonous for fishes.

Control of water pollution:
We have seen that the two sources of water pollution are: Sewage and industrial wastes. They should be removed from water before it is put to use.

Treatment of sewage:
1. Sewage must be churned by machines so that the large pieces may break into smaller ones and may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle and the water flowing down is relatively pure.

2. Water must be sterilized with the help of chlorination. It kills microbes of sewage fungus as well as some pathogens, spores or cytes. Chlorination is very essential particularly in rainy season.

3. Treatment of water with alum, lime etc, also helps in its purifications.

Treatment of industrial wastes:
The treatment of industrial waste depends upon the nature of the pollutants present. In order to ascertain it, the pH of the medium is first determined and the wastes is then neutralised with the help of suitable acid or alkalis. (MPBoardSolutions.com) The chemical substances present in the industrial wastes dissolve in water can be precipitated by suitable chemical reaction and removed later on from water quite recently. Photocatalysed and ion – exchanges have been developed for the treatment of industrial wastes.

Question 16.
What is soil pollution? Write down methods of prevention of soil pollution?
Answer:
Soil pollution:
Change in physical and chemical property of soil due to humans and natural cause is known as soil pollution.
Soil pollution can be prevented by the following methods:

1. Solid and unusable substances like iron, copper, glass, polythene, etc. should not be hurried under soil.
2. Banning cutting of forest and uncontrolled grazing. Crop cycle to be adopted. Suitable arrangement for irrigation to be made. Control on flood and appropriate use of chemical fertilizers and insecticides.
3. Minimum use of chemical fertilizers, insecticides and pesticides.
4. Special attention on recycling of solid waste on melting.
5. Emphasis on the use of cow – dung and human excreta as bio – gas.
6. Biological insecticides to be used.
7. Using closed mines for disposal off waste.
8. Methods of soil erosion to be checked.
9. Soil management to be adopted.
10. Encouraging the use of biofertilizers.

Question 17.
Write the effects of depletion of ozone layer?
Answer:
Sunlight contains ultraviolet radiations and ozone layer present in the atmosphere prevents ultraviolet radiation to reach the earth surface. Continuous depletion of ozone layer cannot prevent ultraviolet radiations from reaching the earth surface and following disadvantages will occur.

1. Intensity of sunlight will increase and temperature of the environment will become intolerable.
2. Increase in skin diseases.
3. Skin cancer becomes common.
4. Immune system will turn weak.
5. Germination and development of seed slows down.

MP Board Class 11 Chemistry Important Questions

ΨMP Board Class 12th Maths Important Questions Chapter 2 Inverse Trigonometric Functions

Inverse Trigonometric Functions Important Questions

Inverse Trigonometric Functions Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
If sin^-1x – cos ^-1 x = \(\frac { \pi }{ 6 } \), then the value of x is equal to:
(a) \(\frac{1}{2}\)
(b) \(\frac { \sqrt { 3 } }{ 2 } \)
(c) \(\frac{-1}{2}\)
(d) None of these
Answer:
(a) \(\frac{1}{2}\)

Question 2.
If tan^-13 + tan^-1 8, then the value of x is equal to:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -3\pi }{ 4 } \)
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 3.
tan^-1 \(\frac{x}{y}\) + tan^-1 \(\frac{x-y}{x+y}\) is equal to:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -3\pi }{ 4 } \)
Answer:
(c) \(\frac { \pi }{ 4 } \)

Question 4.
The value of 2 tan^-1 {cosec (tan^-1 x ) – tan (cot^-1 x)} is equal to:
(a) cot^-1x
(b) cot^-1\(\frac{1}{x}\)
(c) tan^-1x
(d) tan^-1\(\frac{1}{x}\)
Answer:
(c) tan^-1x

Question 5.
The value of tan{cos^-1\(\frac { 1 }{ 5\sqrt { 2 } } \) – sin^-1 \(\frac { 4 }{ \sqrt { 17 } } \)} is equal to:
(a) \(\frac { \sqrt { 29 } }{ 3 } \)
(b) \(\frac{29}{3}\)
(c) \(\frac { \sqrt { 3 } }{ 29 } \)
(d) \(\frac{3}{29}\)
Answer:
(d) \(\frac{3}{29}\)

Question 2.
Fill in the blanks:

1. tan^-1(1) + tan^-1(2) + tan^-1 (3) = …………………………..
2. tan^-1(2) – tan^-1 (1) = …………………………..
3. cot^-1 3 + cosec^-1 \(\sqrt { 5 } \) = ……………………………
4. sin(sin^-1 x + 2 cos^-1 x) = ……………………………….
5. If sin^-1(\(\frac { 2a }{ 1+a^{ 2 } } \)) + sin^-1 (\(\frac { 2b }{ 1+b^{ 2 } } \)) = 2 tan^-1 x, then x = ……………………….
6. If tan^-1 \(\frac{1-x}{1+x}\) = \(\frac{1}{2}\) tan^-1 x ( When x > 0), then x = ………………………..
7. tan^-1\(\frac{a-b}{1+ab}\) + tan^-1\(\frac{b-c}{1+bc}\) + tan^-1c = ………………………….

Answer:

1. π
2. tan^-1\(\frac{1}{3}\)
3. \(\frac{π}{4}\)
4. x
5. \(\frac{a+b}{1-ab}\)
6. \(\frac { 1 }{ \sqrt { 3 } } \)
7. tan^-1 (a)

Question 3.
Write True/False:

1. tan^-1x + tan^-1 y = tan^-1\(\frac{x+y}{1-xy}\)
2. cos^-1(-x) = – cos^-1 x
3. sin^-1(3x – 4x³) = sin^-1 \(\frac{x}{3}\)
4. cos^-1 (\(\frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \)) = 2 tan^-1x
5. sin^-1x – sin^-1[xy – \(\sqrt { 1-x^{ 2 } } \) \(\sqrt { 1-y^{ 2 } } \)]

Answer:

1. True
2. False
3. False
4. True
5. False

Question 4.
Match the column:

Answer:

1. (b)
2. (e)
3. (f)
4. (a)
5. (c)
6. (d)

Question 5.
Write the answer in one word/sentence:

1. Find the value of tan^-1\(\frac{1}{2}\) + tan^-1\(\frac{3}{2}\)
2. Find the value of tan^-1\(\frac { x }{ \sqrt { 1-x^{ 2 } } } \)
3. Find the value of sin ( 2 sin^-1\(\frac{3}{5}\))
4. Solve the equation: sin^-1\(\frac{x}{5}\) + cosec ^-1\(\frac{5}{4}\) = \(\frac { \pi }{ 2 } \)
5. Write the principal value of cos^-1 (cos \(\frac { 7\pi }{ 6 } \))
6. If cos^-1 (\(\frac{1}{x}\)) = θ, then find the value of tan θ

Answer:

1. tan^-1 8
2. sin^-1x
3. \(\frac{24}{25}\)
4. x = 3
5. \(\frac { 5\pi }{ 6 } \)
6. \(\sqrt { x^{ 2 }-1 } \)

Inverse Trigonometric Functions Very Short Answer Type Questions

Question 1.
Find the principle value of the following?

1. tan^-1[sin(- \(\frac { \pi }{ 2 } \)) ] (CBSE 2014)
2. cot [ \(\frac { \pi }{ 2 } \) – 2 cot^-1 \(\sqrt{3}\) ] (CBSE 2014)
3. tan^-1(- \(\sqrt{3}\) )
4. sec^-1 ( \(\frac{-2}{3}\) \(\sqrt{3}\) ) (NCERT)
5. cosec^-1(2) (NCERT)

Solution:
1. Let, tan^-1[ sin(- \(\frac { \pi }{ 2 } \) ) = θ
⇒ tan^-1 [-sin \(\frac { \pi }{ 2 } \) ] = θ
⇒ tan^-1 (-1) = θ
⇒ tan θ = 1
⇒ tan θ = – tan \(\frac { \pi }{ 4 } \)
⇒ tan θ = tan ( \(\frac { -\pi }{ 4 } \) )
θ = \(\frac { -\pi }{ 4 } \)
∴ The principle value is \(\frac { -\pi }{ 4 } \)

2. cot(\(\frac { -\pi }{ 2 } \) – cot^-1 \(\sqrt{3}\))
Let cot^-1 \(\sqrt{3}\) = θ
⇒ cot θ = \(\sqrt{3}\)
⇒ cot θ = cot \(\frac { \pi }{ 6 } \)
∴ θ = \(\frac { \pi }{ 6 } \)
∴ cot ( \(\frac { \pi }{ 2 } \) – cot^-1\(\sqrt{3}\) ) = cot (\(\frac { \pi }{ 2 } \) – 2 × \(\frac { \pi }{ 6 } \))
= cot ( \(\frac { \pi }{ 2 } \) – \(\frac { \pi }{ 3 } \) )
= cot \(\frac { \pi }{ 6 } \)
= \(\sqrt{3}\)
∴ The principal value is \(\sqrt{3}\)

3. Let tan(- \(\sqrt{3}\)) = θ
⇒ tan θ = – \(\sqrt{3}\)
⇒ tan θ = – tan (\(\frac { \pi }{ 3 } \))
⇒ tan θ = tan (- \(\frac { \pi }{ 3 } \))
⇒ θ = – \(\frac { \pi }{ 3 } \)
Hence the principle value is – \(\frac { \pi }{ 3 } \)

4. Let sec^-1( \(\frac{-2}{3}\) \(\sqrt{3}\) ) = θ
⇒ sec^-1( \(\frac { -2 }{ \sqrt { 3 } } \) ) = θ
⇒ sec θ = \(\frac { -2 }{ \sqrt { 3 } } \)
⇒ sec θ = – sec ( \(\frac { \pi }{ 6 } \) )
⇒ sec θ = sec (π – \(\frac { \pi }{ 6 } \) )
⇒ sec θ = sec \(\frac { 5\pi }{ 6 } \)
∴θ = \(\frac { 5\pi }{ 6 } \)
The principle value is \(\frac { 5\pi }{ 6 } \)

5. Let cosec^-1(2) = θ
⇒ cosec θ = 2
⇒ cosec θ = cosec \(\frac { \pi }{ 6 } \) θ ∈ [- \(\frac { \pi }{ 2 } \), \(\frac { \pi }{ 2 } \) ]
The principle value is \(\frac { \pi }{ 6 } \).

Question 2.
Prove the following:

1. 2 cos^-1(\(\frac{4}{5}\)) = cos^-1( \(\frac{7}{25}\) )
2. 2 sin^-1( \(\frac{5}{13}\) ) = sin^-1( \(\frac{120}{169}\) )
3. 2 sin^-1( \(\frac{3}{5}\) ) = sin^-1( \(\frac{24}{25}\) )

Solution:
1. 2 cos^-1( \(\frac{4}{5}\) ) = cos^-1( \(\frac{7}{25}\) )
Formula 2 cos^-1 x = cos^-1(2x² – 1)
∴ L.H.S = 2 cos^-1( \(\frac{4}{5}\) )
= cos^-1 (2 \(\frac{16}{25}\) – 1)
= cos^-1 ( \(\frac{32}{25}\) – 1)
= cos^-1( \(\frac{32-25}{25}\) )
= cos^-1\(\frac{7}{25}\)
= R.H.S.

2. 2 sin^-1\(\frac{3}{5}\) = sin^-1\(\frac{24}{25}\)
Formula 2 sin^-1(x) = sin^-1(2x\(\sqrt { 1-x^{ 2 } } \))
∴ 2 sin^-1 \(\frac{3}{5}\) = sin^-1[2. \(\frac{3}{5}\) \(\sqrt { 1-\frac { 9 }{ 25 } } \)]
= sin^-1[ \(\frac{6}{5}\) \(\sqrt { \frac { 16 }{ 25 } } \) ]
= sin^-1[ \(\frac{6}{5}\) . \(\frac{4}{5}\) ]
= sin^-1[ \(\frac{24}{25}\) ]
= R.H.S. Proved.

3. 2 sin^-1(\(\frac{5}{13}\)) = sin^-1\(\frac{120}{169}\)
Solve like Q.2(b)

Question 3.
Find the value of tan^-1{2 cos(2 sin^-1\(\frac{1}{2}\)} (CBSE 2013, NCERT)
Solution:
tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\)) ]
= tan^-1[ 2 cos (2 sin^-1 sin \(\frac { \pi }{ 6 } \)) ]
= tan^-1[ 2 cos (2. \(\frac { \pi }{ 6 } \)) ]
= tan^-1[ 2 cos \(\frac { \pi }{ 3 } \) ]
= tan^-1 [ 2. \(\frac{1}{2}\) ]
= tan^-1 1
= \(\frac { \pi }{ 4 } \).

Question 4.
Find the value of sin [ \(\frac { \pi }{ 3 } \) – sin^-1( \(\frac{-1}{2}\) ) ]? [CBSE 2008, 2013]
Solution:
sin[ \(\frac { \pi }{ 3 } \) – sin^-1(\(\frac{-1}{2}\) ) ] = sin^-1 [ \(\frac { \pi }{ 3 } \) – ( – sin^-1\(\frac{1}{2}\) ) ]
= sin^-1 [ \(\frac { \pi }{ 3 } \) + sin^-1 sin\(\frac { \pi }{ 6 } \) ]
= sin^-1 [ \(\frac { \pi }{ 3 } \) + \(\frac { \pi }{ 6 } \) ]
= sin^-1 ( \(\frac { \pi }{ 2 } \) )
= 1.

Question 5.
Prove that:
2 tan^-1\(\frac{1}{5}\) = tan^-1 ( \(\frac{5}{12}\) )
Solution:
2 tan^-1( \(\frac{1}{5}\) ) = tan^-1 ( \(\frac{5}{12}\) )
L.H.S. = 2 tan^-1 ( \(\frac{1}{5}\) )

= tan^-1 [ \(\frac{2×25}{5×24}\) ]
= tan^-1 [ \(\frac{5}{12}\) ]
= R.H.S. Proved.

Question 6.
Find the value of tan [ 2 tan^-1 \(\frac{1}{5}\) – \(\frac { \pi }{ 4 } \) ]?
solution:

= tan tan^-1( \(\frac{-7}{17}\) )
= \(\frac{-7}{17}\).

Question 7.
Prove that: 3 sin^-1 x = sin^-1 (3x – 4x³)? (NCERT, CBSE 2018)
Solution:
Let sin^-1 x = θ
⇒ x = sin θ
We know that sin 3θ = 3 sinθ – 4 sin³ θ
= 3x – 4x³
⇒ 3θ = sin^-1 ( 3x – 4x³)
⇒ 3.sin^-1 x = sin^-1(3x – 4x³). proved.

Question 8.
Prove that: 3 cos^-1 x = cos^-1 (4x³ – 3x)? (NCERT)
Solution:
Let cos^-1 x = cos θ
⇒ x = cos θ
We know that cos 3θ = 4 cos³θ – 3 cos θ
= 4x³ – 3x
⇒ 3θ = cos^-1 (4x³ – 3x)
⇒ 3 cos^-1x = cos^-1 (4x³ – 3x). Proved.

Question 9.
Prove the following:

1. tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\) = \(\frac { \pi }{ 4 } \)
2. cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\)
3. cos^-1 \(\frac{3}{5}\) = sin^-1 \(\frac{4}{5}\)

Solution:
1. tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\) = \(\frac { \pi }{ 4 } \)
L.H.S = tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\)

∴ A = tan^-1 \(\frac{5}{12}\)
\(\frac { \pi }{ 4 } \) = R.H.S. Proved

2. cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\)
Let cos^-1 \(\frac{12}{13}\) = A
\(\frac{12}{13}\) = cos A
sin A = \(\sqrt { 1-cos^{ 2 }A } \) = \(\sqrt { 1-\frac { 144 }{ 169 } } \)
= \(\sqrt { \frac { 25 }{ 169 } } \) = \(\frac{5}{13}\)
tan A = \(\frac { sinA }{ cosA } \) = \(\frac { 5/13 }{ 12/13 } \) = \(\frac{5}{12}\)
A = tan^-1 \(\frac{5}{12}\)
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

3. cos^-1 \(\frac{3}{5}\) = sin^-1 \(\frac{4}{5}\)
Let cos^-1 \(\frac{3}{5}\) = A ……………… (1)
⇒ cos A = \(\frac{3}{5}\)
⇒ sin A = \(\sqrt { 1-cos^{ 2 }A } \)
= \(\sqrt { \frac { 16 }{ 25 } } \) = \(\frac{4}{5}\) ……………… (2)
A = sin^-1 \(\frac{4}{5}\).
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

Question 10.
Prove that:

1. sec^-1 x + cosec^-1 x = \(\frac { \pi }{ 2 } \)
2. sin^-1x + cos^-1x = \(\frac { \pi }{ 2 } \)
3. tan^-1x + cot^-1x = \(\frac { \pi }{ 2 } \)

Solution:
1. sec^-1 x + cosec^-1x = \(\frac { \pi }{ 2 } \)
Let sec ^-1 x = θ
∴x = sec θ
⇒ x = cosec ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cosec ^-1 x = \(\frac { \pi }{ 2 } \) – θ. Proved.

2. sin^-1 x + cos^-1 x = \(\frac { \pi }{ 2 } \)
Let sin^-1 x = θ ……………………. (1)
⇒ x = sin θ
⇒ x = cos ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cos ^-1 x = \(\frac { \pi }{ 2 } \) – θ ………………. (2)
Adding eqns (1) and (2),
sin ^-1 x + cos ^-1 x = θ + \(\frac { \pi }{ 2 } \) – θ
⇒ sin ^-1 x + cos^-1 x = \(\frac { \pi }{ 2 } \) Proved.

3. tan ^-1 x + cot^-1 x = \(\frac { \pi }{ 2 } \)
Let tan ^-1 x = θ
⇒ x = tan θ
⇒ x = cot ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cot ^-1 x = \(\frac { \pi }{ 2 } \) – θ
Adding eqns. (1) and (2),
tan ^-1 x + cot ^-1 x = \(\frac { \pi }{ 2 } \). Proved.

Question 11.
Prove the following:

1. tan^-1 5 – tan^-1 3 = tan^-1 \(\frac{1}{8}\)
2. tan^-1 3 – tan^-1 2 = tan^-1 \(\frac{1}{7}\)
3. tan^-1 7 – tan^-1 5 = tan^-1 3 = tan^-1 \(\frac{1}{18}\)

Solution:
1. tan^-1 5 – tan^-1 3 = tan^-1 \(\frac{1}{8}\)
L.H.S. = tan^-1 5 – tan^-1 3
= tan^-1 \(\frac{5-3}{1+5.3}\) = tan^-1 \(\frac{2}{16}\) = tan^-1 \(\frac{1}{8}\)
= R.H.S. Proved.

2. Solve like Q.No. 11 (A).

3. Solve like Q.No. 11(A).

Question 12.
Prove that:

1. tan^-1 \(\frac{4}{7}\) – tan^-1 \(\frac{1}{5}\) = tan^-1 \(\frac{1}{3}\)
2. tan^-1 \(\frac{1}{2}\) – tan^-1 \(\frac{2}{9}\) = tan^-1 \(\frac{1}{4}\)
3. tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{8}\) = tan^-1 \(\frac{3}{11}\)

Solution:
1. tan^-1 \(\frac{4}{7}\) – tan^-1 \(\frac{1}{5}\) = tan^-1 \(\frac{1}{3}\)

2. Solve like Q.No. 12 (A).

3. Solve like Q.No. 12 (A).

Question 13.
tan^-1 1 + tan^-1 2 + tan^-1 3 = π?
Solution:
L.H.S. = tan^-1 1 + (tan^-1 2 + tan^-1 3)
= tan^-1 (1) + π + tan^-1 ( \(\frac{2+3}{1-2×3}\) ),
[∵ tan^-1 x + tan^-1 y = π + tan^-1 \(\frac{x+y}{1-xy}\), if x > 0, y > 0, xy > 1 Here xy = 6 > 1]
= tan^-1 )1) + π + tan^-1( \(\frac{5}{1-6}\) )
= tan^-1 (1) + π + tan^-1 (-1)
= tan^-1 (1) + π – tan^-1 (1), [∵tan^-1 (-x) = – tan^-1 x]
= π = R.H.S. Proved.

Question 14.
(A) If tan^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
tan^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \)

⇒ \(\frac{k+2}{2k – 1}\) = 1
⇒ k + 2 = 2k – 1
⇒2 + 1 = 2k – k
⇒ k = 3.

(B) If tan ^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
Solve like Q.No. 14 (A).
[Answer: k = -1]

(C) If tan ^-1 ( \(\frac{4}{5}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
Solve like Q.No. 14 (A).
[Answer: k = 9]

Question 15.
Prove that:
tan ^-1 \(\sqrt { x } \) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )?
Solution:
R.H.S = \(\frac{1}{2}\) cos ^-1 ( \(\frac{1-x}{1+x}\) )
Let \(\sqrt { x } \) = tan θ
⇒ x = tan² θ
⇒ \(\frac{1-x}{1+x}\) = \(\frac { 1-tan^{ 2 }\theta }{ 1+tan^{ 2 }\theta } \) = cos 2θ
∴ R.H.S. = \(\frac{1}{2}\) cos ^-1(cos 2θ)
= \(\frac{1}{2}\) × 2θ = θ
= tan^-1 ( \(\sqrt { x } \) ) [∵\(\sqrt { x } \) = tan θ ⇒ tan^-1 ( \(\sqrt { x } \) ) = θ]
= L.H.S. Proved.

Question 16.
Prove that:
sin (cos^-1 x ) = cos (sin^-1 x)?
Solution:
L.H.S. = sin(cos^-1 x)
= sin [ \(\frac { \pi }{ 2 } \) – sin ^-1 x],
[∵ sin^-1 x + cos^-1x = \(\frac { \pi }{ 2 } \) , cos^-1x = \(\frac { \pi }{ 2 } \) – sin^-1 x]
= cos (sin^-1 x), [ ∵sin (90° – θ) = cos θ ]
= R.H.S. Proved.

Question 17.
(A) Prove that:
tan^-1 ( \(\frac{b-c}{1+bc}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c = tan^-1 b?
Solution:
L.H.S = tan^-1 ( \(\frac{b-c}{1+bc}\) ) + tan^-1 \(\frac{c-a}{1+ca}\) + tan^-1 a
= (tan^-1 a – tan^-1 b ) + (tan^-1 b – tan^-1 c) + tan^-1 c
= tan^-1 b – tan^-1 c + tan^-1 c – tan^-1 a + tan^-1 a
= tan^-1 b = R.H.S. Proved.

(B) Prove that:
tan^-1 ( \(\frac{a-b}{1+ab}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c = tan^-1 a?
Solution:
L.H.S = tan^-1 ( \(\frac{a-b}{1+ab}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c
= (tan^-1 a – tan^-1 b) + (tan^-1 b – tan^-1c) + tan^-1 c
= tan^-1 a = R.H.S. Proved.

(C) Prove that:
tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)?
Solution:
tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

= tan^-1 \(\frac{20}{91}\) × \(\frac{91}{90}\) = tan^-1 \(\frac{20}{90}\) = tan^-1 \(\frac{2}{9}\)
= R.H.S. Proved.

Question 18.
Solve the equation:
sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) + sin^-1 \(\frac { 2b }{ 1+a^{ 2 } } \) = 2 tan^-1x?
Solution:
sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) + sin^-1 \(\frac { 2b }{ 1+a^{ 2 } } \) = 2 tan^-1x, (given)
⇒ 2 tan^-1 a + 2 tan^-1 b = 2 tan^-1 x [∵sin^-1 \(\frac { 2x }{ 1+a^{ 2 } } \) = 2 tan^-1 x]
⇒ tan^-1 a + tan^-1 b = tan^-1 x
⇒ tan^-1 \(\frac{a+b}{1-ab}\) = tan^-1 x
∴ x = \(\frac{a+b}{1-ab}\).

Question 19.
solve the equation:
cos^-1 ( \(\frac { 1-a^{ 2 } }{ 1+a^{ 2 } } \) ) – cos^-1 ( \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) ) = 2 tan^-1x?
Solution:
cos^-1 ( \(\frac { 1-a^{ 2 } }{ 1+a^{ 2 } } \) ) – cos^-1 ( \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) ) = 2 tan^-1x, (given)
⇒ 2 tan^-1 a – 2 tan^-1 b = 2 tan^-1 x
⇒ tan^-1a – tan^-1 b = tan^-1 x
⇒ tan^-1 \(\frac{a-b}{1+ab}\) = tan^-1 x
∴ x = \(\frac{a-b}{1+ab}\).

Question 20.
(A) Prove the following:
2 tan^-1 \(\frac{1}{4}\) = tan^-1 \(\frac{8}{15}\)?
Solution:
∵ 2 tan^-1 x = tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) )

L.H.S = tan^-1 \(\frac{16}{2.5}\)
= tan^-1 \(\frac{8}{15}\) = R.H.S. Proved.

(B) Prove the following:
2 tan^-1 \(\frac{1}{2}\) = tan^-1 \(\frac{4}{3}\)?
Solution:
We know that 2 tan^-1 x = tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) )

= tan^-1 ( \(\frac{4}{3}\) )
= R.H.S. Proved.

Question 21.
Write in simplest form
tan^-1 \(\sqrt { \frac { 1-cosx }{ 1+cosx } } \)?
Solution:

Question 22.
Write in simplest form:
cos^-1 \(\sqrt { \frac { 1 }{ 2 } (1+cosx) } \)?
Solution:
cos^-1 \(\sqrt { \frac { 1 }{ 2 } (1+cosx) } \) = cos^-1 \(\sqrt { \frac { 1 }{ 2 } .2cos^{ 2 }\frac { x }{ 2 } } \)
= cos^-1 (cos \(\frac{x}{2}\) ) = \(\frac{x}{2}\).

Question 23.
If tan^-1 a + tan^-1 b + tan^-1 c = \(\frac { \pi }{ 2 } \) then prove that ab + bc + ca = 1?
Solution:
tan^-1 a + tan^-1 b + tan^-1 c = \(\frac { \pi }{ 2 } \) , given
⇒ tan^-1 a + tan^-1 b + tan ^-1 c = tan^-1 a + cot^-1 a, [∵tan^-1 a + cot ^-1 a = \(\frac { \pi }{ 2 } \) ]
⇒ tan^-1 b + tan^-1c = cot^-1 a
⇒ tan^-1 ( \(\frac{b+c}{1+bc}\) ) = \(\frac{1}{a}\)
⇒ ab + ca = 1 – bc
⇒ ab + bc + ca = 1. Proved.

Question 24.
Prove that:
tan^-1 \(\frac{2}{11}\) + cot^-1 \(\frac{24}{7}\) = tan^-1 \(\frac{1}{2}\)?
Solution:
L.H.S. = tan^-1 \(\frac{2}{11}\) + cot^-1 \(\frac{7}{24}\)
= tan^-1 \(\frac{2}{11}\) + tan^-1 \(\frac{7}{24}\)

tan^-1 = \(\frac{125}{250}\) = tan^-1 \(\frac{1}{2}\) = R.H.S.

Question 25.
Prove that:
cos^-1 x = 2 cos^-1 \(\sqrt { \frac { 1+x }{ 2 } } \)?
Solution:
R.H.S. = 2 cos^-1 \(\sqrt { \frac { 1+x }{ 2 } } \)
= 2 cos^-1 \(\sqrt { \frac { 1+cos\theta }{ 2 } } \)

= cos^-1 x
= L.H.S. Proved.

Question 26.
Prove that:
cos^-1 x = 2 tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)?
Solution:
R.H.S. = 2 tan^-1 \(\sqrt { \frac { 1-x }{ 1+x } } \)
= 2 tan^-1 \(\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } } \) (putting x = cos θ)

= 2. \(\frac { \theta }{ 2 } \) = θ = cos^-1 x = L.H.S. Proved.

Question 27.
tan^-1 ( \(\frac{a}{b}\) ) – tan^-1 ( \(\frac{a-b}{a+b}\) ) = \(\frac { \pi }{ 4 } \)?
Solution:
tan^-1 ( \(\frac{a}{b}\) ) – tan^-1 ( \(\frac{a-b}{a+b}\) ) = \(\frac { \pi }{ 4 } \)

⇒ 1 = tan\(\frac { \pi }{ 4 } \)
or tan\(\frac { \pi }{ 4 } \) = 1. Proved.

Inverse Trigonometric Functions Long Answer Type Questions – I

Question 1.
(A) Prove that:
sin^-1 \(\frac { 1 }{ \sqrt { 5 } } \) + sin^-1 \(\frac { 1 }{ \sqrt { 10 } } \) = \(\frac { \pi }{ 4 } \)?
Solution:
Let sin^-1 \(\frac { 1 }{ \sqrt { 5 } } \) = A, sin^-1 \(\frac { 1 }{ \sqrt { 10 } } \) = B
∴ A + B = \(\frac { \pi }{ 4 } \)
⇒ sin (A + B) = sin\(\frac { \pi }{ 4 } \)
⇒ sin (A + B) = sin\(\frac { \pi }{ 4 } \)
⇒ sin A cos B + cos A sin B = \(\frac { 1 }{ \sqrt { 2 } } \)
L.H.S = sin A \(\sqrt { 1-sin^{ 2 }B } \) + \(\sqrt { 1-sin^{ 2 }A } \). sin B
= \(\frac { 1 }{ \sqrt { 5 } } \). \(\sqrt { 1-\frac { 1 }{ 10 } } \) + \(\sqrt { 1-\frac { 1 }{ 5 } } \). \(\frac { 1 }{ \sqrt { 10 } } \)
= \(\frac { 3 }{ \sqrt { 5.\sqrt { 10 } } } \) + \(\frac { 2 }{ \sqrt { 5.\sqrt { 10 } } } \)
= \(\frac { 5 }{ \sqrt { 5.\sqrt { 10 } } } \) = \(\sqrt { \frac { 5 }{ 10 } } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
= R.H.S. Proved.

(B) Solve the following equation:
sin^-1 x + sin^-1 (1 – x) = sin^-1 \(\sqrt { 1-x^{ 2 } } \)?
Solution:
Let sin^-1 x = α ∴ x = sin α
Here α + sin^-1 ( 1 – sin α) = sin^-1 \(\sqrt { 1-sin^{ 2 }\alpha } \)
⇒ α + sin^-1 ( 1 – sin α) = sin^-1 cos α
⇒ α + sin^-1 ( 1 – sin α) = sin^-1. sin ( \(\frac { \pi }{ 2 } \) – α)
⇒ α + sin^-1 ( 1 – sin α) = \(\frac { \pi }{ 2 } \) – α
⇒ sin^-1 ( 1 – sin α) = \(\frac { \pi }{ 2 } \) – 2α
⇒ 1 – sin α = sin ( \(\frac { \pi }{ 2 } \) – 2α)
⇒ 1 – sin α = cos 2α
⇒ 1 – cos 2α = sin α
⇒ 2 sin²α = sin α
⇒ sin α = \(\frac{1}{2}\) ∴α = \(\frac { \pi }{ 6 } \)
or x = \(\frac { \pi }{ 6 } \)

Question 2.

1. tan^-1 \(\frac{x+1}{x}\) – tan^-1 \(\frac{1}{2x+1}\) = \(\frac { \pi }{ 4 } \)?
2. If tan^-1 x + tan^-1 y + tan^-1 z = π then prove that x + y + z = xyz?
3. If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac { \pi }{ 2 } \) then prove that xy + yz + zx = 1?

Solution:
1. tan^-1 \(\frac{x+1}{x}\) – tan^-1 \(\frac{1}{2x+1}\) = \(\frac { \pi }{ 4 } \)

= R.H.S.

2. tan^-1 x + tan^-1 y + tan^-1 z = π
⇒ tan^-1 \(\frac{x+y}{1-xy}\) + tan^-1 z = π

⇒ x + y + z – xyz = 0, [∵ tan π = 0]
∴ x + y + z = xyz. Proved.

3. Solve like Q.2(B), take tan \(\frac { \pi }{ 4 } \) = ∞ = \(\frac{1}{0}\)

Question 3.
Write in simplest form:
tan^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]?
Solution:
tan^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]
Let x = tan θ

Question 4.
(A) Prove the following:
\(\frac{1}{2}\) sin^-1 x = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]?
Solution:
R.H.S = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]

= L.H.S. Proved.

(B) Prove that:
\(\frac{1}{2}\) cot^-1 x = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )?
Solution:
\(\frac{1}{2}\) cot^-1 x = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )
R.H.S = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )
Let x = cos θ
R.H.S = cot^-1 ( \(\sqrt { 1+cot^{ 2 }\theta } \) + cot θ )
= cot^-1 ( \(\sqrt { cosec^{ 2 }\theta } \) + cot θ )
= cot^-1 (cosec θ + cot θ)

= \(\frac{1}{2}\) cot^-1 x
= L.H.S. Proved.

Question 5.
Solve the following equation:
tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )?
Solution:
Given: tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )
⇒ tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )

⇒ 17 x = 6 – 3x²
⇒ 3x² + 17x – 6 = 0
⇒ 3x² + 18x – x – 6 = 0
⇒ 3x (x + 6) – 1 (x + 6) = 0
⇒ (x+6) (3x – 1) = 0
∴ x = – 6, x = \(\frac{1}{3}\)

Question 6.
Prove that cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\) = \(\frac { \pi }{ 2 } \)?
Solution:
L.H.S = cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\),

= \(\frac { \pi }{ 2 } \) = R.H.S Proved.

Question 7.
If cos^-1 x + cos^-1 y + cos^-1 z = π then prove that:
x² + y² + z² + 2xyz = 1?
Solution:
Given: cos^-1 x + cos^-1 z = π
⇒ cos^-1 x + cos^-1 y = π- cos^-1 z

Squaring on both sides
x²y² + z² + 2xyz = (1 – x²) (1 – y²)
⇒ x²y² + z² + 2xyz = 1 – y² – x² + x²y²
⇒ z² + 2xyz = 1 – y² – x²
⇒ x² + y² + z² + 2xyz = 1. Proved.

Question 8.
If sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) – cos^-1 \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) = tan^-1 \(\frac { 2x }{ 1-x^{ 2 } } \) then prove that:
x = \(\frac{a-b}{1+ab}\)?
Solution:


⇒ sin^-1 (sin 2θ) – cos^-1 (cos 2ϕ) = tan^-1 (tan 2Ψ)
⇒ 2θ – 2ϕ = 2Ψ
⇒ θ – ϕ = Ψ
⇒ tan^-1 a – tan^-1 b = tan^-1 x
⇒ tan^-1 ( \(\frac{a-b}{1+ab}\) ) = tan^-1 x
⇒ x = \(\frac{a-b}{1+ab}\). Proved.

Question 9.
Prove the following

Solution:
Let x = cos θ, then θ = cos^-1 x.

\(\frac { \pi }{ 4 } \) – \(\frac{1}{2}\) cos^-1 x.

Question 10.
Write in simplest form:
tan^-1 ( \(\frac { x }{ \sqrt { 1+x^{ 2 }-1 } } \) )?
Solution:
tan^-1 ( \(\frac { x }{ \sqrt { 1+x^{ 2 }-1 } } \) )
Putting x = tan θ, we get

Question 11.
Prove that:
tan^-1 \(\sqrt{x}\) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )? (NCERT)
Solution:
tan^-1 \(\sqrt{x}\) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )
R.H.S = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )

Let tan^-1 \(\sqrt{x}\) = θ
\(\sqrt{x}\) = tan θ
R.H.S. = \(\frac{1}{2}\) cos^-1(cos 2θ)
= \(\frac{1}{2}\). 2θ
= θ
= tan^-1 \(\sqrt{x}\)
= L.H.S. Proved.

Question 12.
Prove that:

Solution:


= tan^-1 (tan 3θ) – tan^-1 (tan 2θ)
= 3θ – 2θ
= θ
= tan^-1 2x
= R.H.S. Proved.

Question 13.
Prove that:
cos^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)?
Solution:
Let cos^-1 \(\frac{4}{5}\) = A
∴ \(\frac{4}{5}\) = cos A
∴ sin A = \(\sqrt { 1-cos^{ 2 }A } \) = \(\sqrt { 1-\frac { 16 }{ 25 } } \) = \(\sqrt { \frac { 9 }{ 25 } } \)
⇒ sin A = \(\frac{3}{5}\)
Let cos^-1 \(\frac{12}{13}\) = B
⇒ \(\frac{12}{13}\) = B
∴ sin B = \(\sqrt { 1-cos^{ 2 }B } \) = \(\sqrt { 1-\frac { 144 }{ 169 } } \) = \(\sqrt { \frac { 25 }{ 169 } } \)
⇒ sin B = \(\frac{5}{13}\)
A + B = cos^-1 \(\frac{33}{65}\)
⇒ cos (A + B) = \(\frac{33}{65}\)
⇒ cos A.cos B – sin A.sin B = \(\frac{33}{65}\)
L.H.S. = \(\frac{4}{5}\). \(\frac{12}{13}\) – \(\frac{3}{5}\). \(\frac{5}{13}\)
= \(\frac{48}{65}\) – \(\frac{15}{65}\)
= \(\frac{33}{65}\)
= R.H.S. Proved.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 8 The d-and f-Block Elements

The d-and f-Block Elements Important Questions

The d-and f-Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
General electronic configuration of transition element is :
(a) (n – 1)d^{1 – 10} ns¹
(b) (n – 1)d¹⁰ns²
(c) (n – 1)d¹d^{1 – 10}ns²
(d) (n – 1)d⁵ns¹.
Answer:
(c) (n – 1)d¹d^{1 – 10}ns²

Question 2.
Reason of Lanthanide contraction is :
(a) Negligible screening effect of f – orbitals
(b) Increasing nuclear charge
(c) Decreasing nuclear charge
(d) Decreasing screening effect.
Answer:
(a) Negligible screening effect of f – orbitals

Question 3.
Chromyl chloride test confirms the presence of :
(a) Cl^–
(b) SO₄^2-
(c) Cr³⁺
(d) Cr³and Cl^–.
Answer:
(a) Cl^–

Question 4.
Formula of Mohr’s salt is :
(a) FeSO₄.7H₂O
(b) FeSO₄.(NH₄)2SO₄.6H₂O
(c) CU(OH)₂.CUCO₃.6H₂O
(d) Fe₂O₃.3H₂O.
Answer:
(b) FeSO₄.(NH₄)2SO₄.6H₂O

Question 5.
The outer electronic configuration of chromium is :
(a) 4s¹, 3d⁵
(b) 4s², 3d⁴
(c) 4s⁰, 3d⁶
(d) 4s²,3d⁵.
Answer:
(a) 4s¹, 3d⁵

Question 6.
The equivalent weight of KMnO₄ is alkaline medium will be :
(a) 31.60
(b) 52.66
(c) 79.00
(d) 158.00.
Answer:
(d) 158.00.

Question 7.
The Lanthanide which is widely used :
(a) Lanthanum
(b) Nobelium
(c) Thorium
(d) Cesium.
Answer:
(d) Cesium.

Question 8.
Electronic configuration of Gadolinium is :
(a) [X_e]4f⁶,5d⁹,6s²
(b) [X_e]4f⁷,5d¹6s²
(c) [X_e]4f³,5d⁵,6s²
(d) [X_e]4f⁶,5d²,6s².
Answer:
(b) [X_e]4f⁷,5d¹6s²

Question 9.
In 3d series which element shows highest oxidation state :
(a) Mn
(b) Fe⁺²
(c) Ni
(d) Cr.
Answer:
(a) Mn

Question 10.
Fe, Co, Ni are magnetic substance of which type : (MP 2018)
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.
Answer:
(b) Ferromagnetic

Question 11.
Number of unpaired electrons in Fe⁺² ion is :
(a) 0
(b) 4
(c) 6
(d) 3.
Answer:
(b) 4

Question 12.
In which of the compounds Mn shows highest oxidation state :
(a) K₂MnO₄
(b) KMnO₄
(c) MnO₂
(d) Mn₃O₄.
Answer:
(b) KMnO₄

Question 13.
The atomic radius and ionic radius of Zr and Hf are similar due to :
(a) Diagonal relationship
(b) Both are present in same group
(c) Lanthanide contraction
(d) Similar chemical properties.
Answer:
(c) Lanthanide contraction

Question 14.
Transition elements are coloured due to :
(a) Paired electron in d – orbital
(b) Paired electron in f – orbital
(c) Unpaired electron in d – orbital
(d) None of these
Answer:
(c) Unpaired electron in d – orbital

Question 15.
Stability of ferric ion is due to :
(a) Half filled d – orbital
(b) Half filled f – orbital
(c) Completely filled d – orbital
(d) Completely filled f – orbital.
Answer:
(a) Half filled d – orbital

Question 2.
Fill in the blanks :

1. Metals Fe, Co, Ni are known as …………………….
2. Ionic size of trivalent cations are ……………………. with increase in atomic numbers.
3. The transition metals having lower oxidation state shows ……………………. nature.
4. K₂Cr₂O₇ is a strong ……………………. agent, which gives nascent oxygen.
5. Zn shows only ……………………. oxidation state.
6. f – block elements are known as ……………………. elements.
7. Transition elements and their compounds act as …………………….
8. General electronic configuration of inner transition element is …………………….
9. Chemical form of Potassium manganate is …………………….
10. d – block elements are also known as …………………….

Answer:

1. Ferrous metals
2. Decreases
3. Basic
4. Oxidising, 3
5. +2
6. Inner transition,
7. Catalyst
8. (n – 2)f^{1 – 14} (n – 1)d^{1 – 2}(n – 1)d^{1 – 2}ns²
9. K₂MnO₄
10. Transitional Elements.

Question 3.
State true or false :

1. Mercury is liquid and its oxidation state is +1 and +2.
2. Higher oxidation state of transition elements are acidic in nature.
3. Lanthanides and Actinides both are transition elements.
4. In all transition elements normal oxidation state is +2.
5. Zn, Cd, Hg represent variable oxidation state.
6. Cu⁺² ion is colourless and diamagnetic.
7. Plutonium used as fuel in nuclear reaction and in formation of atomic bomb.
8. Transition elements easily form interstitial compounds.

Answer:

1. True
2. True
3. False
4. True
5. False
6. False
7. True
8. True.

Question 4.
Match the following :

Answer:

1. (f)
2. (g)
3. (e)
4. (c)
5. (b)
6. (d)
7. (a)

5. Answer in one word / sentence :

1. Which is colourless Cu²⁺or Cu⁺?
2. In a reaction KMnO₄ is replaced by K₂MnO₄ then what will be change in oxidation state of Mn?
3. Which series shows higher oxidation state lanthanides or actinides?
4. Which oxidation state of lanthanum is most stable?
5. Write the equivalent weight of K₂Cr₂O7 in acidic medium.
6. How many unpaired electrons are present in Fe³⁺?
7. Give the name of oxidising agent used in chromyl chloride test.
8. Out of d – block elements, Zn does not show variable valencies, why?
9. Which is the most important oxidation state of Cu?
10. d – block elements can be divided into how many series?
11. What is Lunar caustic?
12. In d – block elements Zn does not exhibit variable oxidation state. Why?
13. What is the alkaline solution of HgCl₂ and KI known as?

Answer:

1. Cu⁺²
2. 1
3. Actinides
4. +3
5. 49
6. 5
7. K₂Cr₂O₇
8. Completely filled ‘d’ orbitals
9. +2
10. 2
11. AgNO₃ (Silver nitrate)
12. Due to fully filled d – orbitals
13. Nessler’s reagent.

The d-and f-Block Elements Very Short Answer Type Questions

Question 1.
Actinide contraction is greater from element to element than lanthanide contraction. Why? (NCERT)
Answer:
This is due to poor shielding effect by 5f electrons in the actinoids than that of 4f electrons in the lanthanoids.

Question 2.
Explain Cu⁺ is colourless while Cu⁺² is coloured.
Answer:
If a transition metal contain unpaired electron, it shows paramagnetism and forms coloured compound. In Cu⁺d – orbital is partially filled (3d⁹) thus Cu⁺ is colourless and diamagnetic while Cu⁺² is coloured and paramagnetic.

Question 3.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 oxidation state? (NCERT)
Answer:
Mn⁺² has stable electronic configuration [Ar]4r⁰3d⁵ and they do not easily change to Mn⁺³, Fe⁺² [Ar] 4s⁰3d⁶ on oxidation forms Fe⁺³ [Ar] 4s⁰3d⁵ a more stable configuration.

Question 4.
What are interstitial compounds? Why are such compounds well known for transition metals? (NCERT)
Answer:
Most of the transition elements form interstitial compounds at high temperature with atoms of non – metallic elements like H,B,C,N, Si etc. Small atoms of these non – metallic elements fit in the interstitial voids of crystal lattice of transition elements. These are called interstitial compounds.

Question 5.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. (NCERT)
Answer:
An alloy is a homogeneous mixture of two or more metals or metals and non – metals. An important alloy contains lanthanoid metal is mischmetal which contains 50% Cerium and 25 % Lanthanum, with small amounts of Nd (Neodymium) and Pr (Praseodymium). It is used in Mg – based alloy to produce bullets, shell and lighter flints.

Question 6.
Ti²⁺, V²⁺ and Cr²⁺ are strong reducing agents. Why?
Answer:
For Ti²⁺, V²⁺ and Cr²⁺, values of M²⁺/M is negative which justify that they are strongly reducing.

Question 7.
Write the unit of magnetic moment.
Answer:
Bohr Magneton (BM).

The d-and f-Block Elements Short Answer Type Questions

Question 1.
Wnat is lanthanoid contraction? What are the consequents of lanthanoid contraction? (NCERT)
Answer:
Interesting feature of the atomic size of lanthanides is that on moving down the group steady decrease in atomic size is observed. The shape of f – orbital is in such a way that its shielding effect is minimum, there fore on addition of extra electron in f – subshell only attractive force increases. The steady decrease (contraction) in size of fourteen lanthanide elements (La³⁺1.06 Å to Lu³⁺ 0.8 Å) by a value of about 0.2Å is known as lanthanide contraction.

Reason:
1. The new electrons in lanthanides instead of going to outermost shell enters (n – 2)f – suborbital as a result of which force of attraction increases between electron and nucleus due to which atom or ion contracts.

2. Electron entering in (n – 2)f – suborbital have negligible or zero shielding effect over electrons present in the last orbit. In addition the shape of f – suborbital is not favourable for the shielding effect of electrons. Thus, lanthanide contraction occur.

Consequences of lanthanide contraction :
1. Change in the properties of lanthanides : Due to lanthanide contraction, little change occurs in the properties of lanthanides. So it is very difficult to obtain them in pure state.

2. Influence over the properties of other elements : Lanthanide contraction have an important influence over the element present before and after it e.g., there is difference in properties of Ti and Zr while Zr and Hf have similar properties.

Question 2.
What are Transition elements? They show metallic character. Why?
Answer:
Elements whose atoms in their ground state or ions in their common oxidation states have incomplete or partially filled d – orbitals are called transitional elements. They are in group 2 to 13. Example : Fe, Ni, Co, etc.
General formula : (n – 1)d^{1 – 10}ns^{1 – 2}

Metallic character of an element depends on its tendency to form cation by loosing one or more electrons from its atom. All transitional elements are metals because they contain one or two electrons in their outermost shell which can be easily lost due to low ionisation energy. Thus, they are metallic in nature.

Question 3.
Why do transition metals exhibit variable oxidation states?
Answer:
Transition metals exhibit variable valency because the energy subshell (n – 1 )d and ns are very close. Thus, possibility to lose electrons from ns subshell as well as from (n -1 )d subshell is very much if there are unpaired electrons. So oxidation states of these metals may increase. In these elements Mn shows maximum variable valencies.

Question 4.
Transition elements form alloy easily. Explain.
Answer:
It is the homogeneous mixture of two or more metals or metals with non – metals. Alloys are made to confer the property of metals. Transition elements have great tendency to form alloys because these elements have similar atomic size and can mutually substitute their positions in their crystal lattice. Alloys are comparatively hard and have higher m.p. than the elements from which they are made.

Question 5.
The radius of Fe²⁺ ion is smaller than the radius of Mn²⁺ion, why?
Answer:
The atomic number of Fe (26) is more than the atomic number of Mn (25). Due to higher value of atomic number, iron nucleus contains more protons. Hence the force of attraction between the nucleus and the electrons of outermost orbit is more. Due to strong attractive force of the nucleus the electron cloud is pulled inwards which results in smaller size of Fe²⁺ ion as compared to Mn²⁺ ion.

Question 6.

1. Transition metals possess the ability to form complex compounds. Explain.
2. Zn, Cd and Hg do not show the properties of Transition elements.
3. Why is Ti known as a wonder metal?

Answer:
1. Cause of formation of complex compounds by Transition metals :

• Small size of ions of these elements and high nuclear charge due to which these ions attract ligands.
• They possess vacant d – orbitals in order to accomodate the electron pair donated by ligand.

2. Elements in which (n – 1) d – orbital is partially filled are known as Transition elements Whereas in Zn [3d¹⁰4s²], in Cd [4d¹⁰ 5s²] and in Hg [5d¹⁰6s²] state is found. Therefore, these do not show the properties of Transition elements.

3. Titanium is a shining white metal. It is extended strong (harder than steel), has high m.p. Good conductor of electric current resistant to corrosion and light metal. Due to all these qualities, it is called wonder metal.

Question 7.

TiO₂ is white whereas TiCl₃ is violet, why?
In first transitional series paramagnetism increases till Cr then it starts de – creasing. Why?

Answer:
1. In TiO₂, Ti is in +4 oxidation state (3d⁰4s⁰) having a vacant rf-orbital hence there is no d – d transition and it is white. On the other hand, in TiCl₃, Ti is in +3 oxidation state (3d¹4s⁰) having one unpaired electron in its 3d – orbital, hence it is coloured.

2. In first transitional series, the number of unpaired electrons till Cr (3d⁵) increases and then due to pairing the number of unpaired electrons decreases. Thus, due to this at first paramagnetismjacreases till Cr and then it decreases.

Question 8.
Write five differences between Lanthanide and Actinide.
Answer:
Differences between Lanthanides and Actinides Elements :
Lanthanides Elements:

• Lanthanides show oxidation state of + 3 mainly and +2 and +4 in few compounds.
• Tendency to form complex compound is low.
• Lanthanide compounds are less basic than actinide compounds.
• These do not form oxo – ions.
• Except promethium all are non – radioactive elements.
• Last electron enters in 4f – subshell.

Actinides Elements:

• Actinides show + 3 oxidation state together with + 4, + 5 and +6 in all compounds.
• Tendency to form complex compund is more than lanthanides.
• Actinide compounds are more basic.
• Actinides form oxo – ions as UO₂⁺, NpO⁺, PuO₂⁺, etc.
• All actinides are radioactive.
• Last electron enters in 5f – subshell.

Question 9.
Write chromyl chloride test with equation.
Answer:
Chromyl Chloride Test:
1. When a metal chloride is heated with solid potassium dichromate and cone. H₂SO₄ orange coloured vapours of chromyl chloride are formed.
K₂Cr₂O₇ + 6H₂SO₄ + 4KCl → 2CrO₂Cl₂ ↑ + 6KHSO₄ + 3H₂O

2. When these fumes are passed in sodium hydroxide solution, yellow solution of sodium chromate is obtained. When lead acetate is added to it in presence of acetic acid yellow precipitate of lead chromate is obtained.

Question 10.
Write any five main differences between d – and f – Block elements.
Answer:
Differences between d and f – Block Elements :
d – Block Elements:

• Two shells n and (n – 1) are incomplete.
• Last electron enters the d – orbital of penultimate shell.
• d – block elements are normally called Transitional element.
• d – block elements are available in nature.
• These elements exhibit variable oxidation state.
• These elements are stable.

f – Block Elements:

• Three shells n, (n – 1) and (n – 2) are incomplete.
• Last electron enters the orbital of antipenultimate (n – 2) shell.
• f – block elements are normally called Inner Transitional element.
• f – block elements are very rare. Therefore, they are known as Rare Earth elements.
• These elements also exhibit variable oxidation state.
• These elements are less stable and many are radioactive.

Question 11.
Explain giving reasons: (NCERT)

1. Transition metals and many of their compounds show paramagnetic behaviour.
2. The enthalpies of atomisation of the transition metals are high.
3. The transition metals generally form coloured compounds.
4. Transition metals and their many compounds act as good catalyst.

Answer:
1. Paramagnetic substance is one which is attracted by magnetic field. It arises due to presence of unpaired electron in atom, ion or molecule. Most of the transition elements and compounds are paramagnetic in nature. This is due to fact that transition elements involve partially filled d – subshell and their atom and ion contain unpaired electron.

2. Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

3. The colour of transitional metal ions is due to partially filled (n – 1 )d orbitals. In transitional metal ions which contain unpaired d electrons, transition of electrons takes place from one d – orbital to another d – orbital. During this transition it absorbs some radiation of visible light and reflects the remaining radiation in the form of coloured light. Thus, the colour of the ion is complementary to the colour absorbed by it.
For example:
[Cu(H₂O)₆]²⁺ ion appears blue because it absorbs the red colour of the visible light for electron promotion and reflects its complementary blue colour.
Colour of some ions:

4. Transition elements act as good catalysts in chemical reaction in the hydrogenation of Ni metal, in contact process of manufacture of SO₃, Pt and in manufacture of NH₃ by Haber process Fe acts as catalyst. In the method of preparation of O₂ by heating KClO₃, MnO₂ acts as catalyst.

Question 12.
How would you account for the following : (NCERT)

1. Of the d⁴ species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.
2. Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
3. The d¹ configuration is very unstable in ions.

Answer:
1. Cr⁺² is reducing in nature as its configuration changes from d⁴ to d³ (A stable configuration having half filled t_2g orbitals). On the other hand, Mn⁺³ is oxidising in nature as the configuration changes from d⁴ to d⁵ (A stable configuration having half filled t_2gto e orbitals)

2. Strong ligands force Cobalt (II) to lose One more electron from 3d – subshell and thereby induce d₂sp₃ – hybridisation.

3. The ions with dl configuration try to lose the only electron on d – subshell in order to acquire stable inert gas configuration.

Question 13.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (NCERT)

1. Electronic configuration
2. Atomic and ionic sizes
3. Oxidation state and
4. Chemical roactivity.

Answer:
Differences between Lanthanoids and Actinoids :
Lanthanoids:

• Differentiating or last electrons enter in 4f – sub – shell of (n – 2) orbit.
• These elements come after lanthanum so these are called lanthanoids.
• Common oxidation state is +3, other oxidation states are +2 and +4 also.
• Atomic or ionic radius decreases gradually and this is called lanthanide contraction.
• Lanthanoids have smaller tendency to form complexes.
• Lanthanoids do not form oxo – ions.
• Compounds of lanthanoids exhibit less basic in nature.
• Lanthanoids are not radioactive except Promethium.
• Except Pm, other lanthanoids are present in nature in abundance comparatively more than iodine.

Actinoids:

• Differentiating or last electrons enter in 5f – sub – shell of (n – 2) orbit.
• These elements come after actinium so these are called actinoids.
• Common oxidation state in actinoids is also +3 but other oxidation states are higher, example  +4, +5, +6 and +7.
• Atomic or ionic radius also decreases gradually and steadily and this is qallejj actinoid contraction.
• Actinoids have comparatively higher tendency of complex formation.
• Oxo – ions are formed. example UO₂⁺,PuO₂⁺, UO⁺, etc.
• Compounds of actinoids are more basic in nature.
• All the actinoids are radioactive.
• Most of these are not found in nature and are artificially prepared.

Question 14.
What are Inner Transition elements? (NCERT)
Answer:
These are the elements which contain (n-2)f and (n-1)d incomplete orbitals or in which electron enter in the antipenultimate (two energy levels below the outermost orbital) orbital. These are so called because these are found within the transition elements. There are two types of inner transition elements :
(i) Lanthanides series :
The 14 elements after Lanthanum (La₅₇)
i.e., 58^Ce – 71^Lu are called lanthanides.

(ii) Actinide series : 14 elements after Actinide (AC₈₉) i.e., Th₉₀ to LW₁₀₃.

The d-and f-Block Elements Long Answer Type Questions

Question 1.
Describe the preparation of K₂Cr₂O₇ from chromite ore and explain the reactions of K₂Cr₂O₇ with acidic FeSO₄, KI and H₂S.
Answer:
(A) Preparation:
It is prepared from chromite ore or ferrochrome of chrome iron FeCr₂O₄ (FeO.Cr₂O₃). Different steps involved in the process are as follows :

1. Preparation of sodium chromate:
The ore is finely powdered, mixed with sodium carbonate and quick lime and then roasted (heated to redness) in a reverberatory furnace in presence of excess of air when sodium chromate (yellow in colour) is formed with the evolution of CO₂. Quick lime is added to keep the mass porous and thus facilitates oxidation.

The roasted mass is the extracted with water when sodium chromate dissolves com-pletely leaving behind ferric oxide.

2. Conversion of sodium chromate to sodium dichromate:
Sodium chromate is extracted with water and acidified with sulphuric acid to get sodium dichromate.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O
On concentration the less soluble sodium sulphate Na₂SO₄.10H₂O crystallizes out. This is filtered hot and allowed to cool when sodium dichromate Na₂Cr₂O₇.2H₂O separates on standing.

3. Conversion of sodium dichromatic into potassium dichromate:
Hot concentrated solution of sodium dichromate is treated with requisite amount of potassium chloride when potassium dichromate being less soluble crystallizes out on cooling.
Na₂Cr2O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

(B) Reaction of K₂Cr₂O₇ with acidic FeSO₄, KI and H₂S :
(i) It oxidizes ferrous sulphate to ferric sulphate.

(ii) It liberates I₂ from KI.

These reactions are used in the estimation of iodine and ferrous ion in volumetric an-alysis.
(iii) It oxidizes SO₂ to sulphuric acid.

(iv) It oxidizes H₂S to sulphur.

Question 2.
Explain the oxidizing property of KMnO₄ in acidic, neutral and alkaline medium giving two examples each.
Answer:
KMnO₄ acts as strong oxidizing agent in acidic, neutral and alkaline medium. In acidic medium : It oxidizes in presence of dilute H₂SO₄ and get reduced.
2KMnO₄ +3H₂SO₄ → K₂SO₄ +2MnSO₄ +3H₂O + 5[O]
Example:
1. It oxidizes ferrous salt into ferric salt.

2. It oxidizes oxalate to CO₂ :
2KMnO₄ + 3H₂SO₄ + 5C₂H₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂

3. It oxidizes iodide ion to iodine :
2KMnO₄+10KI+8H₂SO₄ → 6K₂SO₄ + 2MnSO₄+ 8H₂O + 5I₂

4. It oxidizes nitrites to nitrates :
2KMnO₄ + 3H₂SO₄ + 5NaNO₂ → 2MnSO₄ + K₂SO₄ + 5NaNO₃ + 3H₂O

In neutral medium:
In this medium, the reaction begins with neutral ethylene glycol but this does not give neutral reaction because KOH formed in the reaction makes basic in nature.
2KMnO₄ + H₂O → 2KOH + 2MnO₂ + 3[O]
Example:
1. It oxidizes manganous sulphate to manganese dioxide.
2KMnO₄ + 3MnSO₄ + 2H₂O → 5MnO₂ + K₂SO₄ + 2H₂SO₄

2. It oxidizes hydrogen sulphide to sulphur.
2KMnO₄ + 4H₂S → 2MnS + K₂SO₄ + 4H₂O + S

In alkaline medium:
In alkaline medium, reduces to MnO₂ and gives 3 nascent oxygen.
Example:
1. It oxidizes ethylene to ethylene glycol

2. It oxidizes iodide to iodate.

KMnO₄ gives more number of nascent oxygen in acidic medium than in alkaline medium due to which it acts as stronger oxidizing agent in acidic medium.

Question 3.
Describe the preparation of KMnO₄ from pyrolusite and explain its oxidising properties in acidic, basic and neutral medium by suitable example. (MP 2009 Set B, 17)
Answer:
Preparation:
Potassium permanganate is prepared from manganese dioxide. On a large scale, it is prepared from the mineral pyrolusite. The process involves the following steps:

1. Conversion of MnO₂ into potassium manganate:
The finely powdered pyrolusite mineral is fused with potassium carbonate or potassium hydroxide in presence of atmospheric oxygen or an oxidising agent such as potassium nitrate or potassium chlorate. The fused mass turns green due to the formation of potassium manganate.

The fused mass turns green due to the formation of potassium manganate.
2MnO₂ + 2K₂CO₃ + O₂ → 2K₂MnO₄ + 2CO₂
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
MnO₂ + 2KOH + KNO₃ → K₂MnO₄ + KNO₂ + H₂O
3MnO₂ + 6KOH + KClO₃ → 3K₂MnO₄ + KCl + 3H₂O

2. Oxidation of potassium manganate into potassium permanganate :
(i) Chemical oxidation:
The fused mass is extracted with water and the solution is filtered. The green solution is then converted to potassium permanganate by bubbling carbon dioxide, chlorine or oxygen through it.
3K₂MnO₄ + 2CO₂ → 2KMnO₄ + MnO₂ ↓ + 2K₂CO₃
2K₂MnO₄+ Cl₂ → 2KMnO₄ + 2KCl
2K₂MnO₄ + H₂O + O₃ → 2KMnO₄ + 2KOH + O₂
The purple solution of potassium permanganate thus obtained is concentrated when it deposits dark purple, needle like crystals having a metallic lustre.

(ii) Electrolytic oxidation: Nowadays, it is largely manufactured by the electrolytic oxidation of the manganate. The manganate solution is electrolysed between iron electrodes separated by diaphragm. The oxygen evolved at the anode converts manganate to permanganate.
2K₂MnO₄ + H₂O + [O] → 2KMnO₄ + 2KOH
MnO₄^2- + e^– Oxidation (At anode)
2K⁺ + 2e^–→ 2K Reduction (At cathode)
2K + 2H₂O →  2KOH + H₂

After the oxidation is completed, the solution is filtered and evaporated under controlled condition to obtain the crystals of potassium permanganate.
(i) Acidified KMnO₄ solution oxidizes Fe(II) ions to Fe(III) ions i.e. ferrous ions to ferric ions.

(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

MP Board Class 12th Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 10 s – Block Elements

s – Block Elements Important Questions

s – Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Plaster of Paris is :
(a) (CaSO₄)₂.H₂O
(b) CaSO₄.2H₂O
(c) CaSO₄.H₂O
(d) CaSO₄
Answer:
(a) (CaSO₄)₂.H₂O

Question 2.
Lowest melting point compound:
(a) LiCl
(b)NaCl
(c) KCl
(d) RbCl
Answer:
(a) LiCl

Question 3.
Active constituent of Bleaching powder:
(a) CaOCl₂
(b) Ca(OCl)Cl
(c) Ca(O₂Cl₂)
(d) CaCl₂O₂
Answer:
(b) Ca(OCl)Cl

Question 4.
Hydration energy of Mg²⁺ ion will be more than:
(a) Al³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺
Answer:
(b) Na⁺

Question 5.
Which magnetic property is present in alkaline earth metals:
(a) Diamagnetic
(b) Paramagnetic
(c) Ferro – magnetic
(d) Anti – magnetic
Answer:
(a) Diamagnetic

Question 6.
Solubility of which sulphate is the least:
(a) BaSO₄
(b) MgSO₄
(c) SrSO₄
(d) CaSO₄
Answer:
(a) BaSO₄

Question 7.
Compounds of which element are mainly covalent:
(a) Ba
(b) Sr
(c) Ca
(d) Be
Answer:
(d) Be

Question 8.
Important ore of magnesium is:
(a) Malachite
(b) Kaesiterite
(c) Camallite
(d) Galena.
Answer:
(c) Camallite

Question 2.
Fill in the blanks:

1. Radius of Ca²⁺ ion is less than K⁺ because of ……………………………
2. On heating Rb(ICl₂) decomposes to form ………………………….. and ………………………….
3. Be(OH)₂ is soluble in both acid and base because it is of ……………………… nature.
4. Lithium resembles ………………………… element of group 2.
5. Sodium metal gives blue colour in liquid ammonial this is because of ………………………………
6. Potassium forms three oxides ………………………. and …………………………..

Answer:

1. High positive charge
2. RbCl, ICI
3. Amphoteric
4. Mg
5. e^–(NH₃)₂
6. K₂O, K₂O₂, KO₂

Question 3.
Answer in one word/sentence:

1. Sodium metals are kept in kerosene oil. Why?
2. What is lithopone?
3. Which element is present in teeth and bones?
4. What is Caustic soda?
5. What is Sorrel cement?
6. Write formula of Plaster of Paris?

Answer:

1. Highly reactive
2. BaSO₄ and ZnS
3. Calcium
4. NaOH
5. Mixture of MgCl₂ and MgO
6. (CaSO₄)2H₂O

s – Block Elements Very Short Answer Type Questions

Question 1.
Arrange the alkali metals in the increasing order of their reactivity?
Answer:
The reactivity of alkali metals increases from top to bottom.
Be < Mg < Ca < Sr < Ba < Ra.

Question 2.
Write the name of two ores of lithium with formula?
Answer:
Ores of lithium are:

1. Spodumene LiAl(SiO₃)₂
2. Lepidolite Li₂Al₂(SiO₃)₃.F(OH)₂.

Question 3.
Write the name of two ores of magnesium?
Answer:

1. Camolite
2. Dolomite.

Question 4.
Why alkali metals are kept in kerosene oil?
Answer:
Due to high reactivity.

Question 5.
What is the formula of global salt?
Answer:
Na₂SO₄.10H₂O.

Question 6.
What is Lithophone?
Answer:
BaSO₄ + ZnS.

Question 7.
What is the formula of hydrolite?
Answer:
CaH₂.

Question 8.
Carnolite is ore of which metal?
Answer:
Magnesium (Mg).

Question 9.
Which alkali metal show radioactivity?
Answer:
Radium.

Question 10.
Which compound is useful in purification of air in aircrafts?
Answer:
Potassium superoxide.

Question 11.
Name one mineral in which Ca and Mg both are present?
Answer:
Name: Dolomite, Formula: MgCO₃.CaCO₃.

Question 12.
Which flux is used for the removal of acidic impurities in metallic processes?
Answer:

1. Limestone CaCO₃
2. Magnesite MgCO₃.

Question 13.
Formula of Nitrolium is?
Answer:
CaCN₂ and C.

Question 14.
Which forms curtain of smoke?
Answer:
SiCl₄.

Question 15.
What is used to join the fractured bone and make statues?
Answer:
Plaster of paris.

Question 16.
What is the order of stability of carbonates of alkali metals?
Answer:
BeCO₃ < MgCO₃ < CaCO₃ < SrCO₃.

Question 17.
Write the configuration of alkali metals?
Answer:
ns^1-2.

Question 18.
What is the formation of Apsum and Gypsum salt?
Answer:
Magnesium sulphate (MgSO₄.2H₂O) and Calcium sulphate (CaSO₄.2H₂O).

s – Block Elements Short Answer Type Questions – I

Question 1.
Why alkali metal do not found in free state?
Answer:
Alkali metals are highly reactive and electropositive because the value of ionization energy is very low due to larger size thus alkali metals easily donate electron and form positive ion. (MPBoardSolutions.com) These positive ion easily combine with oxygen, moisture and other electronegative element present in nature and form ionic compound.

Question 2.
Explain, why sodium is less reactive than potassium?
Answer:
The ionization enthalpy of sodium is higher than that of potassium. Therefore, sodium lose electron less readily as compared to potassium. Hence, sodium is less reactive than potassium.

Question 3.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium bicarbonate being more soluble than sodium bicarbonate and does not get precipitated when CO₂ is passed through a concentrated solution of KCl with NH₃.

Question 4.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
Lithium is less electropositive than sodium and therefore, carbonates of lithium is less stable than that of sodium. Li₂CO₃ is not so stable to heat and therefore, decomposes at lower temperature. (MPBoardSolutions.com) This is because lithium being veiy small in size polarizes a large. CO₃^2- ion leading to the formation of Li₂O and CO₂. On the other hand, Na₂CO₃ is very stable and decomposes at higher.

Question 5.
I – A and II – A group is called s – block elements. Why?
Answer:
I – A and II – A group elements have their electronic configuration of ns¹ and ns². So the last electron fill in 5 – block so the I – A and II – A group is called 5 – block elements.

Question 6.
What is Soral cement and write its uses?
Answer:
When MgCl₂ solution reacts with MgO and product is formed as MgCl₂.2MgO. nH₂O. It is a white paste and called as Soral cement.
Uses:

1. It is used to fill the teeth cavity.
2. They use in porcelain in point.

Question 7.
Find the oxidation state (O.S.) of Na in Na₂O₂?
Answer:
Let the O.S. of Na in Na₂O₂ = x
One peroxide bond (Na – O – O) is present in Na₂O₂ where the O.S. of O = -1

So, in Na₂O₂ the O.S. of Na = +1.

Question 8.
How calcium sulphate is prepared? Write its uses?
Answer:
In the lab, calcium sulphate is formed by Ca oxide, carbonate, chloride. They react with dil. H₂SO₄ to form calcium sulphate.
CaO + H₂SO₄ → CaSO₄ + H₂
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 9.
What is Gypsum? How plaster of Paris is formed by gypsum?
Answer:
Calcium sulphate CaSO₄.2H₂O is called gypsum. Gypsum is heated at 120° – 130°C and 3 parts of water molecule are released and plaster of Paris is formed.

The plaster of Paris again absorb the water molecule and again get converted to calcium sulphate (Gypsum).

Question 10.
Why during the formation of quick lime the temperature of furnance is not kept more than 1000°C? Explain with equation?
Answer:
During the formation of quick, lime the temperature is maintained at 1000°C because at high temperature the clay present as impurity in the limestone combines with lime producing fusible silicate which fill the pores of lime. Due to this reason slaking of lime become very difficult.
1000°C.
CaO + SiO₂ \(\underrightarrow { 1000^{ \circ }C } \) CaSiO_3.

Question 11.
Why sodium kept in kerosene oil?
Answer:
Alkali metals are very reactive specially against electronegative elements like oxygen, moisture and CO₂. These are oxidized quickly from oxide and hydroxide due to their reactivity. Sodium is kept in kerosene oil to prevent oxidation on exposure to air.
4Na + O₂ → 2Na₂O
2Na + H₂O → Na₂O + H₂
Na₂O + H₂O → 2NaOH.

Question 12.
Write the formula of lime water. What happens when CO₂ passes through it?
Answer:
The formula of lime water is Ca(OH)₂. Due to the flow of the CO₂ gas in Ca(OH)₂ it becomes milky due to formation of a white precipitate of calcium carbonate. (MPBoardSolutions.com) On passing excess of CO₂, milkyness disappears and calcium bicarbonate is formed which is soluble in water.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂.

Question 13.
Why K₂CO₃ is not prepared by Solvay method?
Answer:
Potassium carbonate cannot be prepared by Solvay method because the potassium salt analogous to NaHCO₃ is KHCO₃ which is much soluble and hence, cannot be obtained by crystallization.

Question 14.
Which metal is used in photochemical cell and why?
Answer:
In photochemical cell potassium and caesium metals are used, as their ionization energy is very low.

Question 15.
Why the extraction of sodium from sodium chloride cannot be done by general reducing agents?
Answer:
Sodium is a strong reducing agent. In electrochemical series, it is present at top position. Due to new availability of strong reducing agent than sodium, it cannot be reduced by normal reducing agents. It can only be reduced by electrolytes.

Question 16.
Why Li and Be have the tendency to form covalent compounds? Explain?
Answer:
Due to small size of Li and Be atoms and their high ionization energy, the electrons of the valence shell are firmly bound to their nuclei. The polarizing power of their ions is also high due to high charge density, hence Li and Be form covalent compounds.

Question 17.
Write the Lewis structure of O^–₂ ion and write the O.N. of each O atom. In this ion what is the average oxidation state of O?
Answer:
Lewis structure of O^–₂ ion =
Oxygen without charge have 6 electrons. So, its O.S. = 0, but oxygen with -1 oxidation state have 7 electrons so its O.S. = —1.
Oxidation state of each oxygen atom = \(\frac{-1}{2}\)
O₂^– = 2x = -1
x = \(\frac{-1}{2}\)

Question 18.
In Solvay’s process, can be obtain sodium carbonate by direct reaction of ammonium carbonate and sodium chloride?
Answer:
No, because the reaction between ammonium carbonate and NaCl.
(NH₄)₂CO₃ + 2NaCl ⇄ Na₂CO₃ + 2NH₄Cl
As the products obtained are highly soluble the equilibrium will not shift to,yards forward direction. This is the reason why NaCO₃ cannot be prepared by reaction of (NH₄)₂CO₃ and NaCl in Solvay’s process.

Question 19.
AH compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents? Explain?
Answer:
The size of LT ion is small and thus, it has high polarizing power. This brings covalent character in lithium compounds. Due to covalent character, Li compounds are soluble in organic solvents.

Question 20.
Why are potassium and caesium rather than lithium used n photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light, easily emit elec arm but lithium does nN. Therefore, K and Cs rather than Li are used in photoelectric cells.

Question 21.
Beryllium chloride (BeCl₂) produces smoke when kept in air. Why?
Answer:
Normally air contains moisture, thus beryllium halide hydrolyses in water and releases HCl due to which it produces smoke in air.
BeCl₂ + 2H₄O → Be(OH)₂ + 2HCl.

Question 22.
On moving downward in the first group the hardness of the elements increases. Why?
Answer:
In the first group on moving downwards along with the increase in size cf the elements their density also increases and the force of attraction between its atoms increases due to which there is an increase in their hardness.

s – Block Elements Short Answer Type Questions – II

Question 1.
Why s – block and p – block elements are called representative elements?
Answer:
The elements present in the s and p-block are called normal or representative elements.

1. The elements of group 1 and 2 constitute the s – block of the periodic table. These are known as s – block elements because the last electron in them enters the s – orbital of the valence shell.
2. The elements of p group of the periodic table these are known as p – block elements because the last electron in them enters the p-orbital of the valence shell.

Question 2.
Why alkaline metals are strong or reducing agents?
Answer:
Due to less ionization energy of alkali metals. They have tendency to loose electron (Get oxidized) and form positive ion (M⁺) also alkali metals have negative value of standard reduction potential. (MPBoardSolutions.com) Therefore, alkali metals are good reducing agent.

Question 3.
Why are BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₂ and BaSO₄ are insoluble?
Answer:
The hydration energy of SO₄ of alkaline earth metals decreases down the group. The high hydration energy of Be²⁺ and Mg²⁺ diminishes the lattice energy due to this their SO₂ are soluble in water. But in case of Ca²⁺, Sr²⁺ and Ba²⁺ the hydration energy is low due to this the SO₄ are insoluble in water.

Question 4.
Why the reducing power of lithium is high in solution?
Answer:
Electrode potential is a measure of the tendency of an element to loose electron in aqueous solution. It may depend on the following three factors:

With the small size of its ion, Li has the highest hydration enthalpy. However, ionization enthalpy of Li is highest among alkali metals but hydration (MPBoardSolutions.com) enthalpy predominates over I.P. Therefore, Li is the strongest reducing agent in aqueous solution.

Question 5.
Explain, why can alkali and alkaline earth metals not be obtained by chemical reduction method?
Answer:
The s – block elements themselves are good reducing agents therefore, reducing agent better than s – block elements are not available. Therefore, the chlorides, oxides etc. of s – block elements cannot be reduced to obtain metal.

Question 6.
Explain why alkaline metals form M⁺ cation not M⁺² type of cation?
Answer:
There is one electron in the valence shell of alkali metals. The ionization energy is very low due to bigger size. So they can easily donate electron and can form M⁺ cation. (MPBoardSolutions.com) In the M⁺ state the electronic configuration becomes similar to nobel gases and stable so in this state they become unreactive and the ionization energy become high. So they did not form M²⁺ ion.

Question 7.
In alkali metals, which metal is strongest reducing agent. Why?
Answer:
The reducing character increases from sodium to caesium. However, lithium is the strongest reducing agent among all the alkali metals. Inspite of its highest I.P. This is because of extensive hydration of Li⁺ ions and large amount of energy released during hydration more than compensates the higher I.P. value of Lithium.

Question 8.
Why alkali metals are not found in free state in nature? Or, Why alkali metals always form ionic compounds?
Answer:
Due to bigger size of alkali metals, the ionization energy is very low. So they can easily donate electron and form positive ion. So due to positively charge and highly reactive nature they combine with the (MPBoardSolutions.com) electronegative elements in the nature like moisture, CO₂ etc. and form ionic compounds. Therefore, they cannot be found in free state in nature.

Question 9.
Why alkali metal give flame test?
Answer:
The ionization energy of alkali metal is very low. When these elements or their compounds are heated in Bunsen flame electron present in valency shell absorb energy and easily goes to higher energy level. When these excited electron comes to ground state they emit energy in the form of radiation in visible region and give characteristic colour to flame.

Question 10.
Why Be and Mg do not give flame test?
Answer:
Beryllium and Magnesium atoms in comparison to other alkaline earth metals are comparatively smaller and their ionization energy are very high. (MPBoardSolutions.com) Hence, the energy of the flame is not sufficient to excite their electrons to higher energy levels. These elements therefore do not give any colour in Bunsen flame temperature.

Question 11.
When an alkali metal dissolves in liquid ammonia, the solution acquires different colours? Explain the reason for this type colour change?
Answer:
All alkali metals dissolve in liquid ammonia giving highly conducting deep blue solutions.
M + (x + y )NH₃ M⁺ (NH₃) → x + e^– (NH₃ )_y
When ordinary light falls on these ammoniated electrons. They get excited and jump to higher energy levels by absorbing energy corresponding to red region of the visible light. (MPBoardSolutions.com) As a result transmitted light is blue which imparts blue colour to the solution. However, when the concentration increases the ammoniated metal ion may get bound by free electrons and colour becomes copper bronze.

Question 12.
Why Be and Mg do not give flame test but other metals give? Why?
Answer:
Except beryllium and magnesium all the alkaline earth metals impart characteristic colours to Bunsen flame. Due to small size of Be and Mg atom, the energy required to excite the valency electrons is very high which is not obtained in Bunsen flame. That is why Be and Mg do not impart any colour to flame.

Question 13.
Give two uses of each:

1. Caustic soda
2. Sodium carbonate
3. Quick lime.

Answer:
1. Uses of Caustic soda:

• In soap and paper industry.
• For mercerization of cotton thread.

2. Uses of Sodium carbonate:

• Washing soda is used for washing clothes.
• Removes permanent hardness of water.

3. Uses of Quick lime:

• For making statues, floor, buildings in the form of marble.
• For manufacture of lime, cement, glass and washing soda.

Question 14.
Alkali metals form blue coloured solution when dissolved in ammonia, which is a strong electrolyte. Give reason with equation?
Answer:
Alkali metals dissolve in liquor ammonia to form deep blue coloured solution of high electrical conductivity.
M + (x+y) NH₃ → [M(NH₃)_x] + [e(NH₃)_y]^–
The blue colour of the solution is due to ammoniated electrons positive ion and electrons are responsible for the conductivity.

Question 15.
Na is alkaline or Na₂O? Clarify the statement?
Answer:
Monoxide of all alkali metals are alkaline and form strongly alkaline solution in water.

Na₂O is alkaline because it reacts with water to form NaOH. Na also react with water to form NaOH but it first forms Na₂O and then NaOH, thus Na₂O us alkaline not Na.
4Na + 2H₂O → 2Na₂O + 2H₂
Na₂O + H₂O → 2NaOH

Question 16.
Compare Alkali metals and Alkaline earth metals on the basis of following points:

1. Reaction of heat on carbonate
2. Reaction with nitrogen
3. Solubility of sulphates in water.

Answer:
Comparison of Alkali metals and Alkaline earth metals:

Question 17.
Hydroxides and Carbonates of (Sodium and Potassium) Alkali metals are completely soluble in water whereas that of (Magnesium and Calcium) Alkaline earth metals are partially soluble?
Answer:
All alkali metal carbonates are soluble in water because the value of their Lattice energy is less than their hydration energy.
Lattice energy < Hydration energy (Compound soluble)
Alkaline earth metal carbonates are insoluble because the value of their Lattice energy is more than their hydration energy.
Lattice energy < Hydration energy (Compound insoluble)
∆H_solution = ∆H_{Hydrogen energy} + ∆H_{Lattice energy}

Question 18.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of smallest size among alkali metals, Li⁺ can polarize water molecule more easily than the other alkali metal ions and hence get attached to lithium salts as water of crystallization.

Question 19.
Why is LiF almost soluble in water whereas LiCI soluble not only in water but also in acetone?
Answer:
To make compound water soluble, its lattice energy should be low and hydration energy should be high. Lattice energy of LiCl is less than that of LiF. The difference in these two energies for LiCl and LiF is 31 and 14 kJ/mol respectively. (MPBoardSolutions.com) The difference is larger for LiCl, which is soluble in water. Moreover, lattice energy of LiF is much higher than that of LiCl. It makes LiF sparingly soluble in water. Moreover, LiCl is largely covalent, so it is soluble in organic solvent such as acetone.

Question 20.
If the alkali metals are kept open in the air, after sometime the metallic brightness is lost. Why?
Answer:
On exposure to moist air alkali metals soon get covered with a thick crust of their oxide, hydroxides and carbonate hence, their surface get tarnished. For this reason these metals are stored under kerosene and which prevents them from coming in contact with air and moisture.
4M + O₂ → 2M₂O
M₂O + H₂O → 2MOH
2MOH + CO₂ → M₂CO₃ + H₂O
Here, M is any alkali metal.

Question 21.
Among LiCl and RbCl which will ionize more. Why?
Answer:
Among LiCI and RbCl, RbCl is more reactive because LiCl is slightly covalent in nature due to which it is soluble in organic solvent like pyridine and alcohol.

Question 22.
Among alkali metals and alkaline earth metals whose carbonates are soluble in water? Or, The carbonates of alkali metals are souble in water but that of alkaline metals are insoluble?
Answer:
Carbonates of alkali metals are soluble in water because its hydration energy is greater than lattice energy. Whereas alkaline earth metals are smaller in size and have high density, so its hydration energy is lower than lattice energy. So the carbonates of alkaline earth metals are insoluble in water.

Question 23.
Give the differences between the 8eCl₂ and other chlorides of alkaline earth metals?
Answer:
1. Anhydrous halide are deliquescent and they absorb moisture or water forming hydrated salt.
Example: MgCl₂.6H₂O, CaCl₂.6H₂O, BaCl₂.2H₂O, etc.

2. BeCl₂ fumes on hydrolysis in moist air.
BeCl₂ + 2H₂O → Be(OH)₂ + 2HCl

3. BeCl₂ has different structure in solid and in vapour state. In solid state, it exists as polymeric chain in which each Be atom is surrounded by four chlorine atom. Two of the chlorine atom are covalently bonded while, other two are bonded through co – ordinate bond. In vapour state, at a temperature above 1200 K it has linear monomeric structure with zero dipole moment. Below 1200 K it exist as a dimer.

4. Except BeCl₂ and MgCl₂ other metal chloride give characteristic colour to the flame.

5. Anhydrous CaCl₂ has strong affinity for water therefore, it is used as dehydrating agent.

Question 24.
Why solubility of BaSO₄ is less than CaSO₄?
Answer:
The solubility of sulphates decreases from top to bottom as lattice energy is same. But as we move top to bottom the atomic size increases and hydration energy decreases so solubility decreases.

Question 25.
Why ionization energy of Be is more than B?

These orbitals of Be is full – filled so this element is stable but orbitals of B is not full – filled (Completely), so this element is not stable.

Question 26.
How sodium carbonate is formed by soda process? Write its principle?
Answer:
Solvay method or Ammonia soda process: It has replaced Le – Blanck process and is most commonly employed.
Principle:
In this process, concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammoniacal sodium chloride. On passing CO₂ gas to this solution, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate.

Question 27.
How will you prepare

1. Sodium bicarbonate
2. Sodium hydroxide
3. Sodium silicate from sodium carbonate?

Answer:
1. In sodium carbonate (Aqueous solution) pass CO₂ gas and sodium bicarbonate is formed (White precipitate).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

2. Sodium carbonate is boiled with limewater then sodium hydroxide is formed.
Na₂CO₃ + Ca(OH)₂ → 2Na0H + CaCO₃

3. Sodium carbonate is treated with silica to form sodium silicate.

Question 28.
What is Baking soda? Write its method of preparation, properties and uses?
Answer:
Sodium Bicarbonate (NaHCO₃):
It is also known as sodium hydrogen carbonate or baking soda.

Method of preparation:
Sodium bicarbonate is obtained as an intermediate in the manufacture of sodium carbonate by Solvay’s process. It can be also obtained by passing carbon dioxide through aqueous solution of sodium carbonate.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

Physical property:
It is a white, crystalline solid, slightly soluble in water.

Chemical properties:
1. Effect of heat:
When heated to 373 K it decomposes into Na₂CO₃ and CO₂ is set free.

2. Action of water:
Sodium bicarbonate hydrolyses when dissolved in water. Its aqueous solution is alkaline.

Uses:

1. As a source of carbon dioxide in fire extinguisher.
2. It acts as mild antiseptic for skin infection.
3. As a component of baking powder.
4. As a digestive powder to remove acidity of stomach.

Question 29.
Describe the Le – Blanck method of preparation of sodium carbonate with chemical equation? How Solvay method is better than this?
Answer:
Le – BIanck method:
Sodium chloride when heated with cone. H₂SO₄ produces sulphates and HCl gas. Mixture of sodium sulphate, calcium carbonate and coke on heating provide sodium carbonate and calcium sulphide. This mixture is called blanck ash. (MPBoardSolutions.com) Now in this mixture water is added and filtered. Gas is removed as residue while sodium carbonate being soluble goes into filtrate. Evaporation of filtrate provide solid sodium carbonate.
2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Na₂SO₄ + 4C → Na₂S + 4CO
Na₂S + CaCO₃ → Na₂CO₃ + CaS

Advantages of Solvay method:

1. It is cheap
2. Pure Na₂CO₃ is formed
3. No harmful smoke is obtained
4. In the middle of reaction NaHCO₃ is formed which is a useful compound.

s – Block Elements Long Answer Type Questions – I

Question 1.
Describe the manufacture of caustic soda by Nelson cell on the following point: Labelled diagram, Chemical reactions.

Answer:
It consists of perforated steel tube Graphite anode lined inside with asbestos act as cathode. It is fitted with NaCl solution (brine) and is suspended in the steel tank. Here asbestos lining separates cathode from anode.

One passing current electrolysis of NaCl takes place. (MPBoardSolutions.com) Chlorine is liberated at anode and escapes out. Sodium ions passed through asbestos and is liberated at cathode they reacts with steam for caustic soda.


At anode:

At cathode:
2Na⁺ + 2e^– → 2Na (Reduction)
2Na + 2H²O → 2NaOH + H₂

Question 2.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into ceils. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Answer:
The element is sodium. This imparts yellow colour to the flame. Na⁺ ions participate in the transmission of nerve signals and transport of sugars and amino acids to cell. It forms a monoxide (Na₂O) and a peroxide (Na₂O₂).

The reason for flame colouration is, ionisation energy of Na is low. (MPBoardSolutions.com) Therefore, when Na metal or its salts is heated in Bunsen flame, its valence shell electron is excited to higher energy levels by absorption of energy. When the excited electron returns to the ground state, it emits extra energy in the yellow region of electromagnetic spectrum. Therefore, Na imparts yellow colour to the flame.

Question 3.
Write the diagonal relationship between Be and Al?
Answer:
Diagonal relationship between Beryllium and Aluminium:
Beryllium resembles aluminium of group 3 in the following properties:

1. Both beryllium and aluminium form covalent compounds.
2. Both metals are passive towards reaction with concentrated HN03, because they are covered with a layer of oxide on their surface.
3. Both the metals are of weak electropositive nature.
4. Both do not form hydride quickly.
5. Carbides of both the metals react with water to form methane.

Be₂C + 2H₂O → 2BeO + CH₄
Al₄C₃ + 6H₂O → 2AlCl₃ + 3CH₄

6. Oxides of both are soluble in water and form hydroxides which are amphoteric in nature and react with acid and base to form salt.

BeO + 2HCl → BeCl₂ + H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
BeO + 2NaOH → Na₂BeO₂ + H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

7. Both Be and Al do not give colour to the flame.

Question 4.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). (MPBoardSolutions.com) If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkyness disappears in the last step?
Answer:
The compound (A) is quick lime, CaO. This combines with water and forms calcium hydroxides Ca(OH)₂.

When CO₂ is passed through the solution having Ca(OH)₂, the solution turns milky due to formation of calcium carbonate, CaCO₃.

When excess of CO₂ is passed into milky solution the milkiness disappears due to formation of calcium bicarbonate Ca(HCO₃)₂ which is soluble.

Question 5.
How is lithium different from other members of its group?
Answer:
Anomalous behaviour of Lithium:
Properties of lithium are different than other members of the group due to the following reasons:

1. Size of lithium atom and ion is very small.
2. Due to small size, polarizing power of lithium ion is high, due to w hich compounds possess covalent character.
3. In comparison to other alkali metals its electropositive nature is less and ionization energy is high.

Lithium differ from alkali metals of group I due to following properties:

1. 1. Lithium is comparatively harder than the other alkali metals.
   2. Melting and boiling point of lithium is comparatively high.
   3. Lithium reacts with oxygen to form only normal oxide (Li₂G), whereas other metals form peroxide (M₂O₂) and superoxide (MO₂).
   4. Lithium hydride (LiH) is more stable as compared to hydrides of other metals.
   5. Lithium hydroxide (LiOH) is a weak base and is partially soluble in water, but hydroxides of other metals of the group are more soluble in water.
   6. Lithium forms nitride (Li₃N) with nitrogen, whereas other metals of the group do not form nitride.
   7. Lithium nitrate when heated decomposes to nitrogen dioxide and oxygen.

4LiNO₃ \(\underrightarrow { \Delta } \) 2Li₂O + 4NO₂ + O₂

Whereas sodium nitrate and potassium nitrate when heated strongly form corresponding nitrate with the release of oxygen.

2NaNO₃ \(\underrightarrow { \Delta } \) 2NaNO₂ + O₂

Question 6.
What happens when:

1. Magnesium is burnt in air –
2. Quick lime is heated with silica –
3. Chlorine reacts with slaked lime –
4. Calcium nitrate is heated?

Answer:

s – Block Elements Long Answer Type Questions – II

Question 1.
How is sodium carbonate manufactured by Solvay process?
Answer:
Principle:
In this process, first concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammanica sodium chloride.(MPBoardSolutions.com) On passing CO₂ gas to this, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate. Precipitate of sodium bicarbonate is filter ed which on calcination gives sodium carbonate.
NH₃ + CO₂ + H₂O → NH₄HCO₃
NH₄CO₃ + 2NaCl → NaHCO₃ + NH₄Cl
Na₂HCO₃ → Na₂CO₃ + H₂O + CO₂

CO₂ formed is used agiain for carbonation NH₄Cl formed on treatment with slaked lime gives NH₃ which is used for saturation of brine.
2NH₄CL + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃

The various steps and reactions involved are given below:

1. Saturating tank:
In this tank brine (NaCl) is saturated with ammonia. NaCl solution is introduced from the top while ammonia is introduced from bottom so that ammonium brine is formed and which collects at bottom. Calcium and magnesium salts are present as impurity in sodium chloride solution. They get precipitated as hydroxides by ammonium hydroxide.

2. Filter:
Ammonical brine is filtered to removed the precipitated hydroxides of calcium and magnesium.

3. Cooler:
The filtrate is cooled by passing through condenser. Cooling is necessary because when ammonia dissolves in brine a lot of heat is produced.

4. Carbonating tower:
The cold solution of brine saturated with ammonia is introduced into the carbonating tank from top. Carbonating tower is fitted with a number of compound diaphragms each made of a horizontal iron plate. Following reactions take place here:
2NH₃ + CO₂ + H₂O → (NH₄)₂CO₃
(NH₄)₂CO₃ + 2NaCl → Na₂CO₃ + 2NH₄Cl
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

5. Vacuum filter:
The precipitated sodium bicarbonate along with solution of traces of ammonium carbonate and ammonium chloride is passed through rotatory vacuum filter at the bottom of tower where sodium bicarbonate separates leaving behind mother liquor containing ammonium chloride.

6. Limekiln:
Here limestone is burnt to produce quick lime and carbon dioxide.
CaCO₃ \({ \underrightarrow { \Delta } }\) CaO + CO₂
Formed CaO reacts with water to form slaked lime.
CaO + H₂O → Ca(OH)₂

7. Ammonia recovery tower:
The mother liquor obtained from rotatory filter pump containing ammonium chloride is introduced into the ammonia recovery tank from the top part while the milk of lime is introduced from the top of lower part and steam is introduced into the tower from the bottom, ammonium chloride reacts with milk of lime to produce ammonia.
2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
Sodium bicarbonate obtained from rotatory filters is ignited in a specially constructed cylindrical vessels when sodium carbonate is formed and CO₂ evolved is collected and used again.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Question 2.
What happens when:

1. Reaction of NaOH with Zn or A1
2. Reaction of NaOH with S
3. Reaction of NaOH with halogen
4. Reaction of NaOH with metal oxides
5. Reaction of NaOH with metal salts.

Answer:
1. Zn + 2NaOH → Na₂ZnO₂ + H₂
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

2. 4S + 6NaOH → Na₂S₂O₃ + 2Na₂S + 3H₂O Hypo
8S + 2Na₂S → 2Na₂S₅

3. 2NaOH + Cl₂ → NaCl + NaCIO + H₂O
(Cold) Sodium hypochlorate
6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Warm)

4. ZnO + 2NaOH → Na₂ZnO₂ + H₂O

5. CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂
3NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl
2NaOH + 2 AgNO₃ → Ag₂O + 2NaNO₃ + H₂O.

Question 3.
How Lithium show similarity with Magnesium?
Answer:
In the periodic table some elements differ from elements of the same period but show diagonal similarity with elements of next period.
Example:
Elements of second period show similarities with elements of third period. Like Lithium (second) period with magnesium (third period), Beryllium with A1 and Boron shows similarity with Si.

The relationship between Li and Mg:

1. The Li atomic radius is 1.34 Å and atomic radius of Mg is 1.36 Å .
2. The polar capacity of Li and Mg is equal.
3. Li and Mg is harder elements.
4. The Li and Mg is electronegativity is equal to (1.0 and 1.2).
5. The melting and boiling point of Li and Mg is very high.
6. Li and Mg reacts with N₂ and they form nitrite.
7. Li and Mg reacts with O₂ and they form monoxide.
8. Li and Mg reacts with water to release of hydrogen gas.
9. Li and Mg carbonate is heated and release CO₂ gas.
10. LiOH and Mg(OH)₂ is the weak basicity.

Question 4.
Write the formation of calcium oxide and write its properties and uses?
Answer:
Manufacture:
Commercially, calcium oxide is manufactured by heating limestone.

This process is exothermic and reversible. In order to get good yield of lime, carbon dioxide is to be removed from time to time. (MPBoardSolutions.com) The temperature of the reaction should not exceed 900°C because at higher temperature lime and clay reacts to form fusible silicate.

Furnace used for the manufacture of lime contain two fire boxes, one each on each side. Lime stone is added from the top. It decomposes on reach¬ing down. CO₂ produced is collected and stored in cylinders in liquid state. Lime farmed gets collected at the bottom of the furnace.
Physical properties:

1. It is a white solid.
2. Its melting point is high i.e., 2870K.
3. It is highly stable and it does not decompose in oxy-hydrogen flame but produces a brilliant white light called lime light.

Chemical properties:
1. It reacts with moist air absorbing CO₂ to form Ca(OH)₂ and CaCO₃
CaO + H₂O → Ca(OH)₂ (Slaked lime)
CaO + CO₂ → CaCO₃ (Lime stone)

2. When heated with ammonium salt, ammonia gas evolved.
2NH₄Cl + CaO → CaCl₂ + H₂O + 2NH₃

3. When heated with coke at high temperature, calcium carbide is produced.

4. It forms calcium chloride when chlorine gas is passed over hot and dry calcium oxide.
2CaO + 2Cl₂ → 2CaCl₂ + O₂

5. Calcium oxide is a strong basic oxide. It reacts with acid to form salt and water while, with acidic oxide it forms respective salt.
CaO + 2HCl → CaCl₂ + H₂O
CaO + H₂SO₄ → CaSO₄ + H₂O
CaO + CO₂ → CaCO₃ (Calcium carbonate)
CaO + SiO₂ → CaSiO₃ (Calcium silicate)
CaO + SO₂ → CaSO₃ (Calcium sulphite)
3CaO + P₂O₅ → Ca₃(PO₄)₂ (Calcium phosphate)

6. It forms slaked lime when dissolved in water. It is an exothermic reaction.
CaO + H₂O → Ca(OH)₂ + 15,000 cal.

Uses:

1. For making basic lining in furnaces.
2. For drying alcohol and gases.
3. Lime water is used as laboratory reagent and in medicine.
4. For purification of coal gas and in paper industry.
5. As flux in metallurgical process.
6. For producing lime light.
7. For manufacture of ammonia, sodalime, calcium carbide, cement and glass.

Question 5.
Differentiate between Alkali metals and Alkaline earth metals?
Answer:
Differences between Alkali metals and Alkaline earth metals:
Alkali metals:

1. They show +1 oxidation state.
2. Their hydroxides are strong bases.
3. Their carabonate, sulphate and phosphate compounds of alkaline earth plate are soluble in water.
4. Their ionisation energy is comparatively less.
5. They are melleable, ductile and lustrous.

Alkaline earth metals:

1. These show +2 oxidation state.
2. Their hydroxides are comparatively weaker bases.
3. These compounds of alkaline earth metals are insoluble in water.
4. Their ionization energy is high.
5. These are comparatively less.

Question 6.
Explain the Castner – Kellner cell with diagram to obtained sodium hydroxide?
Answer:
Castner – Kellner Cell:
It consists of rectangular iron tank which is fitted with three slate portion which do not touch the bottom but fit into grooves at bottom. The bottom of cell is covered with mercury and this is brought into circulation with the help of an eccentric wheel. (MPBoardSolutions.com) The outer compartmentA is filled with brine and inner Graphite compartment B with caustic soda solution. Two slot graphite electrode project from the ceiling of the outer compartment of the vessel and act anode. The iron cathode consisting of several rods is filled in middle compartment.

There are exit pipes fitted in compartment and for removal of hydrogen in central compartment. On passing current sodium chloride (brine) solution is electrolyzed in the two outer compartment. Chlorine is liberated at anode. (MPBoardSolutions.com) Sodium amalgam formed is pushed by the help of eccentric wheel into the central compartment. In the central compartment B when caustic soda is electrolyzed OH^– move to the mercury layer which act as anode and after discharge react with the sodium of Na – Hg to form caustic soda with the release of H₂ gas. Simultaneously an equivalent amount of sodium liberated at cathode reacts with water to produce more caustic soda.
2Na – Hg + 2H²O → 2NaOH + 2Hg + H₂↑

The mercury obtained can again be used in the cell. Caustic soda solution is evaporated to obtain solid caustic soda which is fused and cast into flakes or sticks.
Cell reactions are as follows:

1. In outer compartments:
Ionization:
2 NaCl ⇄ 2Na⁺ + 2Cl^–; (Ionization)

At cathode:
2Na⁺ + 2e^– → 2Na (Reduction)
2Na + xHg → Hg_xNa₂ (Sodium amalgam)

At anode:
2Cl^– → 2Cl + 2e^–; (Oxidation)
2Cl → Cl^2-

2. In central compartment:
Ionization:
NaOH ⇄ Na⁺ + OH^–; (Ionization)

At cathode:
2Na – Hg + 2H₂0 2NaOH + 2Hg + H₂
Na⁺ + e^– → Na, (Reduction)
2Na +2H₂O 2NaOH + H_2(g)

At anode:
2OH^– → 20H + 2e^–, (Oxidation)
Hg_xNa₂ + 20H → 2NaOH + xHg

MP Board Class 11 Chemistry Important Questions

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 12 Indian Constitution Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 12 Indian Constitution

MP Board Class 10th Social Science Chapter 12 Text book Exercises

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

Question (i)
The constitution is –
(a) Form of government
(b) Government of country
(c) Documents of rules
(d) Fundamental Rights and laws
Answer:
(c) Documents of rules

Question (ii)
Which of the following is not a speciality of the Indian constitution –
(a) Parliamentary form of government
(b) Federal Government
(c) Free and impartial Judiciary
(d) Unwritten Constitution
Answer:
(d) Unwritten Constitution

Question (iii)
How many Fundamental duties have been indicated in the Constitution –
(a) 5
(b) 14
(c) 18
(d) 10
Answer:
(a) 5

Question 2.
Fill in the blanks: (MP Board 2011, 2013)

1. Chairman of the Constituent Assembly was …………
2. Dr.B.R. Ambedkar was chairman of constitutional ………..
3. The newly drafted Constitution was adopted by the constituent assembly on …………
4. The right of equality is one of the ……………. describe in the Constitution.

Answer:

1. Rajendra Prasad
2. Drafting Committee
3. 26 Nov 1949
4. Rights

Question 3.
Match the coloumn:

Answer:

(a) (iv)
(b) (ii)
(c) (v)
(d) (i)
(e) (iii).

MP Board Class 10th Social Science Chapter 12 Very Short Answer Type Questions

Question 1.
What is a constitution?
Answer:
Constitution may be said to be a document which contains laws, rules and regulations for proper governemnt of a country.

Question 2.
Who is the protector of Fundamental Rights in India?
Answer:
Supreme Court (Judiciary).

Question 3.
When fundamental duties were added in the Constitution?
Answer:
According to the 42th Amendment in 1976 the fundamental duties were added in the Constitution of India.

Question 4.
In whose guidance the Independence Bill was drafted?
Answer:
Lok Manya Bal Gangadhar Tilak.

MP Board Class 10th Social Science Chapter 12 Short Answer Type Questions

Question 1.
What is the importance of Constitution? (MP Board 2010)
Answer:
Constitution is needed for properly running the government and clearly defining the roles and functions of the legislature, judiciary and administration so also for ensuring a balance and coordination amongst them. In the absence of Constitution there is a strong possibility of conflicts between different organs of the Government and at times even a situation of anarchy may arise Fundamental rights and duties of citizens are also specified in the Constitution. It can be said that Constitution is the very base of Government.

Question 2.
Write an introductory note on Constituent Assembly?
Or
Give the introduction of Constitution Assembly? (MP Board 2009)
Answer:
The formation of Constituent Assembly was the result of the ‘Swaraj Bill’ prepared under the directions of Lokmanya Bal Gangadhar Tilak in 1895, the statement of Mahatma Gandhi on 5 January, 1922 that, “Indian Constitution shall be as per the wishes of the people of India”, the National Convention of February 1924 under the Chairmanship of Tej Bahadur Sapru and the demand in the Patna Convention of Swaraj Dal in May 1934.

The members of the Constituent Assembly could not be elected directly by adult franchise therefore as a practical solution provincial Assemblies were utilised as election bodies.

Question 3.
Who were the prominent members of the Constituent Assembly from Madhya Pradesh?
Answer:
Important members of the Constituent Assembly from the existing provinces viz. Central Province a Berar, Central India region (Bhopal, Gwalior, Indore and Rewa) were Pandit Ravishankar Shukla, Seth Govind Das, Hari Vishnu Kamat, Ghanshyam Singh Gucch, Gopikrishan Vijayvargiya, Radhavallabh Vijayvargiya and Thakur Lai Singh.

Question 4.
What do you understand by Directive Principales of the State Policy?
Or
What do you mean by Directive Principles of State Policy? Write. (MP Board 2009)
Answer:
Directive Principles of the State Policy:
In the fourth part of the Constitution, fundamental principles of Governance have been described, these are known as directive principles. The principles are guidelines for political, social and economic programmes for modem democracy. Although these principles cannot be directed by any court to be enforced but they are fundamental in the good Governance of the country. Through these directive principles an attempt has been made to set up a welfare state in India.

Question 5.
What is importance of Preamble in the Constitution?
Answer:
In the Preamble of the Constitution the framers of the Constitution have incorporated the objectives of the Constitution, so also values and ideals. This is called the essence or spirit of the Constitutions, it is the resolve and feelings of the framers of the Constitution.

In the very beginning of the Preamble it has been indicated that the Constitution has been framed as per the wishes of the people and ultimate powers vest in the people. It has been stated that India shall be sovereign, democratic republic. As per the 42nd Constitutional amendment; India was declared as a socialist and secular state.

Question 6.
Explain the meaning of Socialist and Secular. (MP Board 2009, 2013)
Answer:
Socialism and Secular:
By a socialist state is meant that the Indian economy shall be based on socialistic pattern of society. Minimum basic needs of every Indian shall be fulfilled. Socialism as per the Indian conditions shall be adopted. The ideal of secularism has been cherished in the Constitution. It means that the state shall protect the interests of all religious beliefs but will not have any particular religion as state religion. The state shall not discriminate citizens on the basis of religion. Every citizen is free to follow its religion and beliefs.

MP Board Class 10th Social Science Chapter 12 Additional Important Questions

Objective Type Questions

Question 1.
Multiple Choice Questions:
Choose the correct answer from the following

Question (i)
Cabinet Mission Plan came to India in –
(a) 1942
(b) 1944
(c) 1946
(d) 1948.
Answer:
(b) 1944

Question (ii)
Indian Constitution is divided into –
(a) 20 Parts
(b) 21 Parts
(c) 22 Parts
(d) 23 Parts.
Answer:
(c) 22 Parts

Question (iii)
How many articles are there in Indian constitution –
(a) 320
(b) 345
(c) 370
(d) 395
Answer:
(d) 395

Question 1.
Fill in the blanks:

1. The first draft on Constitution was submitted to the President of the Constituent Assembly on …………..
2. The first meeting of Constituent Assembly was held on 9 December………….

Answer:

1. 21 February, 1948
2. 1946

Question 3.
True and False type questions:

1. The demand for Constituent Assembly was related with National Movement.
2. Indian Constitution had 395 Articles at the time of final submission.
3. Supreme Court is the protector of fundamental rights? (MP Board 2009)
4. Dr. Rajendra Prasad was the Chairman of the Drafting Committee of Constitution? (MP Board 2009)

Answer:

1. True
2. True
3. True
4. False

Answer in One – Two Words or One Sentence:

Question 1.
What did the citizens obtain by enjoying fundamental rights?
Answer:
The citizens can live independent and progressive life by enjoying fundamental rights.

Question 2.
How has Abraham Lincoln defined democracy?
Answer:
Abraham Lincoln defined democracy, as that it is government of the people, by the people and for the people.

Question 3.
When was the first meeting of the Constituent Assembly held?
Answer:
First meeting of the Constituent Assembly was held on 9 December, 1946.

Question 4.
Who was elected the Chairman for first meeting of the Constituent Assembly?
Answer:
Dr. Sacchidanand Sinha.

Question 5.
When was new constitation passed and adopted?
Answer:
The new constitution was passed and adopted on 26 November, 1949.

Question 6.
Which plan is the basis of Indian constitution.
Answer:
By 42nd Amendment (1976).

MP Board Class 10th Social Science Chapter 12 Very Short Answer Type Questions

Question 1.
What is Universal adult franchise?
Answer:
The Constitution has given the right to vote to all its adult citizens. By adult franchise is meant that every adult citizen has a right to vote on attainment of a definite age. In the Constitution this right is given to all citizens who attain the age of 18 years, irrespective of their religion, race, caste, sex or place of the person.

Question 2.
What do you mean by a soverign state?
Answer:
The country will decide its own foreign and domestic policies. It is not under any foreign rule and it can have its own policies of behaviour at the international level.

Question 3.
When did the new constitution was adopted?
Answer:
The new Constitution was passed and adopted on 26 November, 1949. Certain provisions of the Constitution in respect of citizenship, election and interim Parliament were adopted with immediate effect and remaining provisions became effective from 26 January, 1950, which was called as the date of promulgation of the Constitution.

Question 4.
What did Dr. B.R Ambedkar say about the Constitution of India?
Answer:
Dr. B.R. Ambedkar said about the Indian Constitution:
It is the bulkiest in the world. Here people of different religions and classes live. Detailed description of all this is given in the constitution of the country.

Question 5.
Explain the right of speech and freedom of expression?
Answer:
The right to freed on grants certain freedoms to the Indians. One such freedom is right to state one’s views and express them. But this freedom is to be exercised within the framework of public order, decency, sovereignty etc.

Question 6.
Explain the right to constitutional remedies?
Answer:
Right to constitutional remedies is one of the important fundamental rights. According to this right, the courts are required to protect the fundamental rights from any encroachment. For this, the courts have the power to issue writs. Habeas Corpus, Prohibition, Quo Warranto, Mandamus, Certiorari.

Question 7.
Why Fundamental Rights are incorporated in the Indian Constitution?
Answer:
The incorporation of fundamental rights in the Constitution is a matter of great significance. This shows that how important and sacred fundamental rights have been considered and how they are thought to be so Fundamental that they have been incorporated in the Constitution.

Question 8.
Why the Indian citizen should follow the Fundamental Duties?
Answer:
The Indian citizen follows and should follow fundamental duties because –

1. Duties constitute part of the constitution and as such possess legal states.
2. They have to enjoy rights and hence should follow duties.

MP Board Class 10th Social Science Chapter 12 Short Answer Type Questions

Question 1.
Explain the right to equality?
Answer:
Right to equality has been explained in Articles 14, 15, 16, 17 and 18. The kind of equality in the constitution is given as under –

1. Equality before law and equal protection of law.
2. Prohibition of discrimination on grounds of caste, creed, colour, sex, region or any one of them.
3. Equal opportunities to all, based only on qualifications.
4. Abolition of untouchability.
5. Abolition of all title except academic and military.

Question 2.
What rights are received from the right against exploitation?
Answer:
The following rights are received from the right against exploitation –

1. Prohibition of bonded labour.
2. Prohibition of Adivasis system.
3. Prohibition of employment of children below the age of 14 in any hazardous beings.
4. Prohibition of trafficing in human beings.

Question 3.
Distinguish between Fundamental Rights and Directive Principles?
Answer:

1. The Fundamental Rights are enforceable while the Directive Principles are not enforceable.
2. The Fundamental Rights are in the form of injustiction on the state while the Directive Principles are in the form of directives to the state.
3. The Fundamental Rights are rights of the citizens the Directive Principles are the duties of the state.
4. The Fundamental Rights lay the foundations of political democracy while the Directive Principles, of economic democracy.
5. The Fundamental Rights are, by and large, negative whereas the Directive Principles are positive in character.

Question 4.
What do you mean by the Federal Form of Government?
Answer:
Federal Form of Government As per the First Schedule of the Constitution, India is a federation of States. Thus federal form of Government has been set up in India. The powers of Government are not Centralized at one place and are divided between the Center and the States and both have independence in their respective jurisdictions.

The Constitution is written and rigid to a considerable extent and it is supreme. The Supreme Court is the protector of the Constitution. Supreme Court has also the powers to interpret the Constitutional provisions and decide Constitutional disputes arising between the Center and the States.

Question 5.
Discribe the salient features of Indian Constitution?
Answer:
Salient Features of Indian Constitution are:

1. Written and the largest Constitution.
2. Mix of rigidity flexibility.
3. Sovereign Republic.
4. Socialist and secular republic.
5. Parliamentary system.
6. Federal system.
7. Free and impartial judiciary.
8. Fundamental rights and basic duties.
9. Directive principles of state policy.
10. Universal adult franchise.

Question 6.
Indian Constitution is the matrix of flexibility and rigidity. Explain?
Answer:
A Constitution is termed as rigid or flexible on the basis of the procedure adopted for its amendment. If the constitution can be adopted by a simple procedure followed for framing simple laws it is termed as flexible but if a special procedure is needed for amendment then it is termed as rigid or inflexible.

There are three procedures of amendments in the Indian Constitution. Certain provisions can be amended by a simple majority, some provisions can be amended by specific majority and certain important ones can only be amended by specific majority and consent of atleast 50 per cent number of states. Thus it is a mix of flexibility and rigidity.

Question 7.
How many types of Constitution can be there?
Answer:
The fathers of Indian Constitution consulted and took useful provisions from Constitutions of several countries. The directive principles have been taken from the Constitution of Ireland. The idea of Fundamental Rights has been taken from the Constitution of United States of America. The concept of Federation of states (Federal form of Government) is taken from Canadian Constitution.

MP Board Class 10th Social Science Chapter 12 Long Answer Type Questions

Question 1.
Describe the characteristics of the Fundamental Rights and Duties in Indian Constitution? (MP Board 2011, 2013)
Or
Describe the fundamental duties of citizen of India? (MP Board 2009)
Or
Describe the fundamental rights and citizen? (MP Board 2009)
Answer:
The Fundamental Rights are as follows:

1. Right to equality.
2. Right to freedom.
3. Right against exploitation.
4. Right to freedom of religion.
5. Cultural and educational rights.

Fundamental Rights and Duties:
For alround development of the citizens, fundamental rights are essential. There is a provision for fundamental righjts for the citizens. In case they are violated, a citizen has the right to approach the High Court or Supreme Court. During the period of Emergency these Fundamental Rights can be suspended.

The provision of Fundamental Duties of the citizens have been added in the Constitution by an amendment (42nd Amendment of the Constitution, 1976). Ten fundamental duties have been specified in the Constitution.

Fundamental Duties:
It shall be the duty of every citizen of India.

1. To abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem.

2. To cherish and follow the noble ideals which inspired our national struggle for freedom.

3. To uphold and protect the sovereignty, unity and integrity of India.

4. To defend the country and render national service when called upon to do so.

5. To promote harmony and spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women.

6. To value and preserve the rich heritage of our composite culture.

7. To protect and improve the natural environment including forest, lakes, rivers and wildlife and to have compassion for living creatures.

8. To develop the scientific temper, humanism and the spirit of inquiry and reform.

9. To safeguard public property and to abjure violence.

10. To strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavor and achievement.

Question 2.
Discuss any four features of the Indian Constitution?
Or
Describe the main features of Indian constitution? (MP Board 2009)
Or
Describe the Parliamentary and Federal System of the Government?(MP Board 2009)
Or
Write any five features of Indian constitution? (MP Board 2009)

Answer:
The constitution of India is the lengthiest constitution. Accordingly it possesses certain important features. Some of these can be discussed as under:

1. Federal Form of Government:
The Constitution does possess all the formal requisites of a federation written constitution, rigidity of the constitution, decentralisation, powerful judiciary. But it can change into a unitary system during emergency.

2. Parliamentary System:
The Present of India is constitutional head whereas the council of Ministers is the real. All the ministers are collectively responsible in the Lok Sabha. The ministers work as a team they swim, and sink together.

3. Abolition of Untouchability:
The constitution abolishes untouchability. Untouchability is declared against law and against the constitution.

4. Universal Adult Franchise:
The universal adult franchise has been guaranteed by the constitution. Every citizen, irrespective of any caste, creed, colour or sex, has the right to vote and participate in administration.

Question 3.
Mention some of the Directive Principles of State Policy?
Answer:

1. The state would prohibit concentration of material and natural resources in the hands of the few.
2. Public assistance to the poor, the aged and the helpless.
3. Right to work and to education.
4. Equal wages for equal work, for both men and women.
5. Organising and empowering the Village Panchayats to make them units of self – government.
6. Promoting cottage and small scale industries in the rural, areas.
7. Enacting prohibition.
8. Prohibiting the slaughter of the cows and those animals which give milk.
9. Organising agriculture on scientific lines and promoting animal husbandry.
10. Providing free and compulsory education to the children below 14 years within a period of ten years.
11. Uniform civil code for the whole country.
12. Separation of executive from judiciary.
13. Protection of monuments of national importance.
14. Promoting world peace and security, cooperation and arbitration.
15. Giving economic and educational assistance to Scheduled Castes and Scheduled Tribes and other weaker sections of the society.

Question 4.
Explain briefly to the Drafting Committee for the preparation of Indian Constitution?
Answer:
Drafting Committee:
Drafting of the Constitution was not an easy task, therefore different sub committees were formed to facilitate drafting of the Constitution. 10 sub – committees were relating to matters of procedure and 8 sub – committees related to matters of facts. Important sub – committees were Rules, Business and Drafting Committee, Steering Committee, Committee on Fundamental Rights Union Powers Committee, Minority Rights Sub-committee, etc.

To give final shape to the reports and suggestions of the above sub – committees a drafting Committee under the chairmanship of Dr. Bhimrao Ambedkar was formed. Other members of this Committee were N. Gopalswami Ayangar, Aladi Krishna Swami Aiyyar, Sayyad Mohammad Sadullah, K.M. Munshi. B.L. Mittra and D.P. Khaitan subsequently B.L. Mittra and D.P. Khaitan were replaced by N. Madhavan Rao and T.T. Krishnamachari. The Draft Constitution was submitted to the President of the Constituent Assembly on 21 February, 1948. Thereafter detailed discussions were held in the Constituent Assembly.

Question 2.
Discuss the cultural base of Indian Constitution.
Answer:
The framers of the Constitution were aware of the fact that the life of the people of a country is closely associated with the culture and rich heritage of the country the identity of a country is not on the basis of political organisation or beliefs but on the basis of the culture, traditions, beliefs and feelings of nationality.

Indian Constitution is not merely a political document but much more than that. It contains the culture and National Values in it. In the manuscript of the Constitution, the rice cultural and national heritage of the country has been depicted through representative pictures. These pictures depict the traditions from Mohan – jodaro and Vedic period to the period of Freedom Movement.

These pictures are of the bull seal of Mohan – jodaro period, Lord Ram returning with Sita in the ‘Pushpak Viman’ after victory of Lanka, preaching of Gita by Sri Krishan to Arjuna, Lord Buddha, Mahavir Swami, Akbar and in the background Mughal architecture, Shivaji, Guru Govind Singh, Tipu Sultan, the historical Dandi March of Mahatma Gandhi, the Himalayas the Indian ocean etc.

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 21 Globalisation Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 21 Globalisation

MP Board Class 10th Social Science Chapter 21 Text book Exercises

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

Question (i)
Globalisation has improved the standard of living of –
(a) Poor class
(b) Higher class
(c) Rural areas
(d) All classes of the society
Answer:
(c) Rural areas

Question (ii)
Which industries are closed due to globalisation –
(a) Large – scale industries
(b) Multinational companies
(c) Small – scale industries
(d) Industries of all type
Answer:
(d) Industries of all type

Question (iii)
The process of globalisation began in India from the year –
(a) 1947
(b) 1951
(c) 1991
(d) 2001.
Answer:
(c) 1991

Question (iv)
The World Trade Organisation was established in –
(a) 1985
(b) 1995
(c) 2001
(d) 2005.
Answer:
(b) 1995

Question (v)
The main basis of globalisation is –
(a) foreign trade
(b) internal trade
(c) agricultural trade
(d) small scale industry.
Answer:
(a) foreign trade

Question 2.
Fill in the blanks: (MP Board 2009)

1. At present the number of reserved industries is …………….
2. Under globalisation the transportation of goods and services is ……………. between various countries.
3. The companies who produce goods in different countries are called ……………. (MP Board 2009)

Answer:

1. Eight
2. Liberalised
3. Mulitnational Companies.

MP Board Class 10th Social Science Chapter 21 Very Short  Answer Type Questions

Question 1.
What was the foreign policy of India prior to the year 1991?
Answer:
Liberalisation and Globalisation.

Question 2.
What are those companies called who produce in more than one country?
Answer:
Multinational Company.

Question 3.
Which consumer class is benefited more by globalisation?
Answer:
Higher Class.

Question 4.
What do you understand by multinational companies?
Answer:
A company whose work is related to production and sale spread all over the world is known as multinational companies.

MP Board Class 10th Social Science Chapter 21  Short  Answer Type Questions

Question 1.
State the meaning of globalisation. Who have been benefited by globalisation?
Answer:
Globalisation is meant the working of whole world together with co – operation and co-ordination in the form of a market. Under the process of globalisation the restrictions on the inflow and outflow of goods and services from one country to another are withdrawn. Thereby the market prices start working freely in whole of the world. As a result the prices in all the countries becomes near about equal. In this way as a result of globalisation all the markets of the world are unified. The upper strata of our society is the most beneficiary of this system.

Question 2.
What is foreign trade?
Answer:
The foreign trade has linked all the countries of the world together. There are several big companies of the world, which are called multinational companies. These companies sell their products in several countries of the world. This is noticeable here that a multinational company is that who produces goods in more than one country. These companies produce goods on large scale and sell these produced goods in several countries.

Question 3.
What is the unification of market?
Answer:
India has adopted the New Economics Policy since 1991. The main objective of this policy is to take advantage of the progress of the world and technical knowledge and thus accelerate the economic development of the country. This policy has opened the way of liberalisation by avoiding the existing administrative restriction. Along with this the efforts has been made to motivate private investment and to attract the foreign capital. It can be said in brief that through the new policy a new chapter has began by linking the India market economy with the economy of the world.

Question 4.
How are the small producers affected by globalisation?
Or
Write the effect of Globalization on the small producers? (MP Board 2009)
Answer:
All people have not been benefited by the globalisation, thus now it is clear that globalisation has affected the Indian economy in both the ways inversely and positively. The reality is this, that globalisation has benefited the industries and business and we have approached in the world market. Several Indian producers have got the form of multinational company.

The development rate of Indian economy has also become more than 9 per cent. Consumers are now getting world – level standardised goods. But in reality the people of all classes have not got the benefit of globalisation equally. Several small and tiny industries have been closed due to competition. The problems like poverty and unemployment prevailing in the developing countries like India have become more complicated.

Along with this due to the influence of developed countries upon. The World Trade Organisation, developing and backward countries are not getting the benefits of globalisation. The new labour – laws are not affecting the labour class favourably. Therefore the efforts are needed.

MP Board Class 10th Social Science Chapter 21 Long Answer Type Questions

Question 1.
What do you understand by globalisation? Explain the causes that motivate the process of globalisation.
Or
What is globalisation? Explain the causes that motivate this process. (MP Board 2009)
Answer:
Globalisation means opening ftp the economy to facilitate its integration with the world economy. In this situation economics are integrated with each other. In case of globalisation it becomes easy to sell goods and services abroad.

Factors Inspiring Globalisation:

1. Expansion of Technical Knowledge:
During last 50 years the technical knowledge has developed rapidly. The transport technology has made it possible to send goods upto distant places in a lower cost. Telecommunication facilities such as internet, mobile phone, fax etc. has made the task of connecting people with °ach other throughout the world very easy. Communication satellite has brought a revolutionary change by expanding these facilities. As a result globalisation has expanded rapidly.

2. Process of Liberalisation:
Till the middle of twentieth century the production was limited mainly within the boundaries of the countries. So many countries imposed strict restrictions to protect the goods produced by them from competition. India also during the decades of 1950 and 1960 had permitted only to import the necessary goods as machineries, fertilizers and petroleum etc. Several industries developed due to this policy and India became self – reliant in several sectors.

3. Expansion of Competition and Market:
Competition has a special importance in capitalist economic system. In this system different producing compaines in order to seek the hold on market take support of competition. For this purpose these companies along with cutting down the prices use advertisements and various medium of convincing and canvassing the buyers.

4. Expansion of Multinational Companies:
Multinational companies play a significant role in linking the distant countries with each other. These companies set up their factories for production purpose in those countries where they get cheaper labour and other means of production and due to this the capacity of these companies increasesto compete.

Question 2.
How do foreign trade helps in unifying the markets of various countries?
Answer:
During the decades of 1970 and 1990 some such changes took place due to which began the process of liberalising foreign trade. For example, dissolution of Union of Socialist Soviet of Russia, economic unification of Europe, emergence of Japan as a major power of the world and economic development of Korea, Singapore and Hongkong. As a result several countries agreed to liberalise the world trade.

This strengthen, the process of liberalization. After the establishment of The World Trade Organisation1 in 1955, almost all the countries of the world have reduced their taxes on imports and have opened the markets of their countries for other countries. So, the process of globalisation has gained speed.

Question 3.
Explain the economic condition of India after the globalisation, and discuss the problems created by gloabaissation?
Or
Explain the main problems which are created by globalisation? (MP Board 2009, 2010)
Answer:
The following aspects are there undertaking The system of globalisation in India:
1. Industrialisation:
The industrialisation in the country has been accelerated due to the industrial policy declared in the year 1991 and the reforms done afterwards. Now number of industries reserved exclusively for public sector is reduced only to three. It means that the private sector has got sufficient opportunities for expansion.

2. Increase in Foreign Investment:
After the globalisation the multinational companies have increased their investment in India. These companies have shown their interest in such investment as cell phones, motorcars, electronic equipments, cold drinks, junk food materials and banking services. New opportunities of employment have been created by these industries and services.

3. Advantage to Indian Companies:
Globalisation has set up several Indian companies as multinational companies. These companies have expanded their activities at world level. For example, Tata Motars, Infosys, Ranbaxy, Asian Paints, Sunderm Fastners can be taken as examples which have now become multinational companies.

The Indian companies, in order to compete with foreign companies, have adopted the latest technology and have raised the standards of their production. There are some companies who have improved their condition by collaborating with multinational companies.

Problems Created by Globalisation:
This is not true to say that India has only benefited by globalisation. The reality is this, the globalisation has also created several problems. Following are these problems.

1. Impact on Small Producers:
Globalisation has adversely affected the several small industries of India. Small industries are not capable to compete with goods produced in foreign. As a result several small industries have closed.

2. Uncertainty of Employment:
Lives of labourers have been greatly affected by globalisation. These days due to growing competition, maximum employers like flexibility in providing employment to labourers. It means that the jobs of labourers are not secure. The factory owners, to minimize the cost, provide temporary employment to the labourers; so that they may not have to pay them salary round the year.

1. Benefit not to all
2. Regional disparities.

MP Board Class 10th Social Science Chapter 21 Additional Important Questions

Objective Type Questions

Question 1.
Multiple Choice Questions:
(Choose the correct answer from the following)

Question (i)
The New Industrial Policy was adopted since –
(a) 1989
(b) 1990
(c) 1991
(d) 1992.
Answer:
(c) 1991

Question (ii)
The main base of globalisation is –
(a) Multinational Companies
(b) Foreign Trade
(c) America
(d) European Union.
Answer:
(b) Foreign Trade

Question (iii)
New economic policy liberated the –
(a) Private sector
(b) Public sector
(c) Industrial sector
(d) Tertiary sector.
Answer:
(a) Private sector

Question (iv)
W.T.O. means –
(a) World Terrorist Organisation
(b) World Transport Organisation
(c) World Trade Organisation.
(d) World Temperate Organisation
Answer:
(c) World Trade Organisation.

Question 2.
Fill in the blanks:

1. World bank is an International ……………… institution that extends financial assistance to member nation for development purposes.
2. Financial Crunch is a situation in which the government ……………… falls drastically short of government expenditure.
3. Privatisation is defined as transfer of ownership and control from the public sector to the ……………… sector.
4. The private sector allowed to establish industries and
   business but subject to ……………… and ………………

Answer:

1. Financial
2. Revenue
3. Private
4. Control regulation

Question 3.
True and False type questions:

1. New Economic Policy or N.E.P. was adopted in 1995.
2. Globalization means decreasing integration between different economics of the world.
3. Privatisation means to bring most of the enterprises of the country under the ownership, control and management of the private sector.
4. Liberalisation means to rid industry and trade of unnecessary restrictions and make them more competitive.

Answer:

1. False
2. False
3. True
4. True.

Question 4.
Match the following:

Answer:

1. (c)
2. (a)
3. (d)
4. (b).

Answer in One – Two Words or One Sentence:

Question 1.
Define World Trade Organisation (WTO)?
Answer:
WTO is the global international organisation dealing with the rules of trade between nations.

Question 2.
When was WTO established?
Answer:
WTO was established on January 1, 1995.

Question 3.
Where is WTO located?
Answer:
WTO is located at Geneva, Switzerland.

Question 4.
What is GATT?
Answer:
GATT (General Agreement on Trade and Tariffs) was the forum for negotiating trade agreement between nations between 1947 – 1994.

Question 5.
When did India become the member of WTO?
Answer:
India became the member of the WTO on January 1, 1995.

Question 6.
Mention two functions of WTO?
Answer:

1. Administering WTO trade agreements.
2. Forum for trade negotiations between countries.

MP Board Class 10th Social Science Chapter 21 Very Short  Answer Type Questions

Question 1.
Define sustainable development.
Answer:
Sustainable development is an economic development where necessities of generations are not compromised by the pleasure of present generation.

Question 2.
Explain the concept of sustainable economic development. Give two points?
Answer:

1. Development should take place without polluting and damaging the environment.
2. Development should be continuous and it should fulfil the needs and aspiration of the present and future generations.

Question 3.
What is the role of WTO? What two benefits will India get by being a member of WTO?
Answer:
WTO is the global trade organisation dealing with the rules of trade between nations. India, as a member of WTO will be benefited in the following ways:

1. Promotion and expansion of trade among countries.
2. (a) Less restrictions on Indian exports.
   (b) Getting technology from developed countries.

Question 4.
Define Liberalisation?
Answer:
Liberalisation means freedom to the private sector to run those activities which were earlier confined to only public sector. Secondly, Private sectors have given many relaxations from the rules and regulations.

Question 5.
What is NRI?
Answer:
NRI (Non-Resident Indian) is an Indian who does not normally live and work in India, but in some other country. He holds Indian citizenship and Indian passport.

MP Board Class 10th Social Science Chapter 21 Short Answer Type Questions

Question 1.
Mention three freedoms given to the private sector industry by liberalisation?
Answer:

1. Industrial licensing has been abolished except for five industries.
2. Number of public sector industries has been reduced from 17 to 3.
3. Freed from regulation like permission for importing materials.

Question 2.
Enlist main objectives of New Economic Policy?
Answer:
Following were the objectives of new economic policy:

1. Liberalisation of the economy.
2. Expansion of private sector.
3. Encouragement of private foreign investment.
4. Modernisation of agriculture.
5. Controlling fiscal deficit.

Question 3.
What do you mean by disinvestment? How far did we succeed in this programme?
Answer:
Disinvestment means selling shares of its performing enterprise by the government. It is one method of privatization. Our programme of privatization through investments has succeeded to some extent only.

Question 4.
Mention the functions of WTO?
Answer:
Functions of World Trade Organisation (WTO):

1. Administering trade agreements between nations.
2. Forum for trade negotiations.
3. Handling trade disputes between nations.
4. Monitoring national trade policy.
5. Technical assistance and training for developing countries and.
6. Cooperation with other international organisation.

Question 5.
How can you justify the presence and impact of globalisation?
Answer:
The presence and impact of globalisation can be justified by the following facts:

1. Development of social consciousness.
2. Fast and quick technological changes.
3. Global form of modern business.
4. Formation of International Monatory Fund (IMF).
5. Foreign collaboration and joint ventures, financial and assistance.
6. Globalisation of marketing through Cable TV network and satellite links and
7. Introduction of internet facilities like e-mail and e- commerce services.

Question 6.
Explain in short meaning of the term ‘Globalization’. Write its main characteristics (features on related points) also?
Answer:
Meaning:
Globalization means integration, unification or integration of domestic economy with the world economy. Characteristics (Features or Related Points):

1. Due to globalisation producers from outside can sell their goods and services in India. We can do the same with our goods and services.
2. Entrepreneurs from other countries can invest in India and the Indian entrepreneur can also do the same in other countries.

Question 7.
Discuss some positive aspects of Globalisation and Liberalisation?
Answer:
Globalisation and Liberalisation have helped in rapid economic development.

1. A large industrial base created; increase in industrial production.
2. Proportion of people living below the poverty line less.
3. Self – sufficient in food.
4. Mobilised its savings.
5. Generated its own resources for development.
6. Large pool of scientists and technically skilled working person.
7. Export – oriented industries.

Question 8
Differentiate between policy of restriction and policy of liberalization?
Answer:
Liberalization:
It means removing unnecessary trade restrictions and making the economy more competitive. New Economic policy liberated the private sector from strict control and licensing.

Restriction:
If restrictions are imposed on economic activities by government policies, it is called the policy of restriction or restrictive policy. No real economy is completely free of restrictions. When these restrictions are removed, it is called the policy of liberalization.

Question 9.
What measures have been taken for globalisation of the economy of India?
Answer:
Free interaction among economies of the world in the field of trade finance, production techlnologies and investment is termed as globalisation of the economy. It encourages foreign trade and institutional investment. Following measures are adopted for it:

1. Devaluation of rupee by 20% in July 1990 – 91.
2. Full convertibility was offered in 1993-94 to encourage exports earnings.
3. Long period trade policy of remove restrictions.
4. Encouragement to open competition.
5. Modification of custom and tariff.

Question 10.
What is sustainable development? What are its important features?
Answer:
The development, which takes care of the needs of present generation without compromising with the needs of future generation is termed as sustainable development.

Important features of sustainable development:

1. Economic growth
2. No or the least compromise with the necessities of the future generation
3. Pollution free economic development
4. No or the least depletion of non – renewable resources and
5. Preservation of environmental and exhaustible resources.

Question 11.
Explain the meaning of World Trade Organisation?
Answer:
World Trade Organisation (WTO):
The global international organisation, working on multilateral trading system, where trade agreemeftts are negotiated and signed by a large majority of world’s trading nations and ratified in their parliaments are known as W.T.O. Its objective is to assist trade flowing smoothly, freely, fairly and predictable among nations. WTO was established on Jan 1, 1995 in Geneva, Switzerland. India also became its member on January 1,1995.

MP Board Class 10th Social Science Chapter 21 Long Answer Type Questions

Question 1.
Review the status of Indian economy before the new economic policy – 1991?
Answer:
1. Self – reliance:
The five – year plans imed at obtaining self – reliance. The dependency upon foreign aid was reduced in fulfilment of this aim. During this period self – reliance in agriculture sector was achieved and vast industrial sector was developed.

2. Foreign Trade:
In the year 1991 imports were kept under control. During this period only necessary goods like machineries, fertilizer and petroleum were mainly imported. To protect the domestic producers from foreign competition the policy of protection was adopted. Therefore, the trade during this period increase slowly. The contribution of India in total world trade in 1951 was near about. 1 per cent which reduced upto 0.6 percent in 1991.

3. National and Per Capita Income:
During the period of 1951 to 1991 the national income increased at the average rate of 4.0 per cent. But due to the rapid growth in population during this period the per capita income increased at a very slow speed.

4. Increase in the Opportunities of Employment:
During this period though the efforts were made to increase the job opportunities even then the problem increased day – by – day. The problem of unemployment became very complicated till 1991.

5. Crisis of Foreign Currency:
India adopted the policy of reducing imports between the years 1951 to 1991. But India needed foreign currency for the import of petroleum products machinery and other necessary goods. India had to take international loan to import these goods. Therefore, India was trapped in the crisis of foreign exchange.

6. Price rise:
During the period of planning India had to face the problem of continuous price rise. After the First Five-Year Plan during the years 1956 to 1991 the rate of inflation in India was between 5 to 6 per cent.

Question 2.
Explain briefly the key features of Indian economy after adoption of the New Economic Policy?
Answer:
Key Features of Indian Economy after adoption of New Economic Policy:
1. Industrial activities which were restricted for public sector were also opened up to private sector except for industries having national importance tike defence, space research etc.

2. Relaxation in regulations tike quota system, industrial licensing, concentration of economic power etc. This was meant to give freedom to the businesses to undertake activities having growth potential.

3. Permission to import raw material was eased to make/it more competitive for businesses to reduce cost and undertake better technology.

4. Pricing and distribution was made free to give businesses free hand to deal with pricing and distribution strategies.

5. Restriction on investement and increase in production capacity was eased to pave the way for industrial growth.

6. The economy was opened to integrate it with world economy in terms of flow of goods and services investment etc.

Question 3.
Explain the process of globalisation?
Or
Give the meaning of globalisation and describe the steps taken in this direction?
Or
Write the main five factors which promote globalization? (MP Board 2009)
Answer:
Process of Globalisation or Steps taken for Globalisation:
1. Raising Foreign Equity Participation:
Prior to July 1999 f foreign equity participation was subjected to lot of approvals, sanctions and constraints. It was restricted to 40%. Now it has been increased to 51% and the approvals have been made routine work.

2. Devaluation of Rupee:
Rupee was devalued by 20% in July 1990 – 91. The devaluation was made to encourage exports and discourage imports. It also aimed at inflow of more foreign capital.

3. Convertibility of Rupee:
The government offered partial convertibility of rupee through the budget of 1992 – 93. Full convertibility was offered in 1993 – 94. Convertibility of rupee was aimed at encouraging export earnings.

4. Long Period Trade Policy:
The government announced foreign trade policy for a period of five years i.e. 1992 – 97. The sole purpose of this policy was liberalisation. It also removed restrictions on external trade.

5. Encouragement to Open Competition:
Exports and imports were left to market forces, government control was minimised.

6. Modification of Customs and Tariff:
In order to build up our competitive strength, customs and tariff policies were modified to promote international trade.

7. Modernisation of the Economy:
The new economic policy accords top priority to modern techniques and technology. It also promotes computers and electronics industries. It has made the Indian industries dynamic. All foreign collaborations concerning higher technology have been (created by the government.

8. Privatisation of the Company:
It means removing strict control over private sector and making them free to take necessary decisions. Now, the new policy tries to expand private sectors.

MP Board Class 12th Chemistry Important Questions Chapter 9 Coordination Compounds

Coordination Compounds Important Questions

Coordination Compounds Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
The correct structural formula of Zeise’s salt is :
(a) K⁺ [PtCl₃(C₂H₄)]^–
(b) K⁺[PtCl₂(η² – C₂H₄)^–]Cl^–
(c) K⁺[PtCl₃(η² – C₂H₄)]
(d) K⁺PtCl₃(η² – C₂H₄)]^–
Answer:
(a) K⁺ [PtCl₃(C₂H₄)]^–

Question 2.
AgCl is soluble in aqueous ammonia due to formation of:
(a) [Ag(NH₃)₂]²⁺
(b) [Ag(NH₄)₂]⁺
(c) [Ag(NH₃)₄]⁺
(d) [Ag(NH₃)₂]⁺.
Answer:
(d) [Ag(NH₃)₂]⁺.

Question 3.
Which of the following gives a white precipitate in aqueous solution with silver nitrate:
(a) [Cr(NH₃)₅Cl](NO₂)₂
(b) [Pt(NH₃)₃Cl₂]
(c) [Pt(CN)₂Cl₂]
(d) [Pt(NH₃)₂]Cl₂.
Answer:
(d) [Pt(NH₃)₂]Cl₂.

Question 4.
Correct nomenclature of Fe₄[Fe(CN)₆]₃ is :
(a) Ferroso ferric cyanide
(b) Ferric ferrous hexacyanate
(c) Iron (III) Hexacyanoferrate
(d) Hexacyanoferrate (III-II).
Answer:
(c) Iron (III) Hexacyanoferrate

Question 5.
In which of the following compounds oxidation state of metal is zero :
(a) [Pt(NH₃)₂Cl₂]
(b) [Cr(CO)₆]
(c) [Cr(NH₃)₃Cl₃]
(d) [Cr(CN)₂Cl₂]
Answer:
(b) [Cr(CO)₆]

Question 6.
Example of dsp² hybridization is :
(a) [Fe(CN)₆]^-3
(b) [Ni(CN)₄]^-2
(c) [Zn(NH₃)₄]⁺²
(d) [FeF₆]^-3
Answer:
(b) [Ni(CN)₄]^-2

Question 7.
Which of the following complex is used as anti-cancer agent:
(a) Trsns[CO(NH₃)₂Cl₃]
(b) cis[Pt(NH₃)₂Cl₂]
(c) cisK₂[PtCl₂Br₂]
(d) Na₂CO₃.
Answer:
(b) cis[Pt(NH₃)₂Cl₂]

Question 8.
Oxidation state of Fe in [Fe(CO)₃] complex is :
(a) -1
(b) +2
(c) + 4
(d) 0
Answer:
(d) 0

Question 9.
Grignard reagent is :
(a) Organometallic compound
(b) Complex compound
(c) Double salt
(d) Neutral compound.
Answer:
(c) Double salt

Question 10.
Structure of complex salt was proposed by :
(a) Berzelius
(b) Werner
(c) Raoult
(d) Faraday.
Answer:
(b) Werner

Question 11.
Mohr’s salt is :
(a) Double salt
(b) Complex salt
(c) Neutral salt
(d) Reagent.
Answer:
(a) Double salt

Question 12.
The formula of nitroprusside is :
(a) Na₄[Fe(CN)₅NO₅]
(b) Na₂[Fe(CN)₅NO]
(c) NaFe[Fe(CN)₆]
(d) Na₂[Fe(CN)₆NO₂].
Answer:
(b) Na₂[Fe(CN)₅NO]

Question 13.
Which of the following is not an organometallic compound : (MP 2018)
(a) C₂H₅MgBr
(b) (C₂H₅)₄Pb
(c) C₂H₅ONa
(d) (CH₃)₄Al.
Answer:
(b) (C₂H₅)₄Pb

Question 14.
Zeigler Natta catalyst is :
(a) (Ph₃P)₃RhCl
(b) K[PtCl₃(C₂H₄)]
(c) [Al₂(C₂H₆)₆]
(d) [Fe(C₂H₅)₂].
Answer:
(a) (Ph₃P)₃RhCl

Question 15.
The I.U.P.A.C. name of [Ni(CO)4] is :
(a) Tetracarbonyl nickelate (0)
(b) Tetracarbonyl nickelate (II)
(c) Tetracarbonyl nickel (0)
(d) Tetracarbonyl nickel (II)
Answer:
(c) Tetracarbonyl nickel (0)

Question 2.
Fill in the blanks :

1. Cis [Pt(NH₃)₂Cl₂] complex is used as an ……………. agent.
2. Haemoglobin is a ……………. compound of iron.
3. Geometrical isomerism is found in both ……………. and ……………. complexes.
4. Oxidation state of Ni in Ni(CO)₄ is …………….
5. Diethyl zinc is a ……………. compound.
6. The correct I.U.P.A.C. name of K₄[Fe(CN)₆] is …………….
7. Oxidation state of Co in [Co (E.D.T.A)] is …………….
8. The formula of dibromo chlorotriaquo chromium (III) is …………….
9. [COF₆]^-3 is a ……………. spin complex.
10. The formula of antiknock organometallic substance is …………….
11. E. D. T. A is ……………. ligand.
12. Example of hexadentate ligand is …………….

Answer:

1. Anti – cancer
2. Complex
3. Tetrahedral, Octahedral
4. Zero
5. Organometallic compound
6. Potassium hexacyano ferrate (II)
7. +3,8. [Cr(H₂O)₃Cl Br₂]
8. High
9. Tetraethyl lead (C₂H₅)₄Pb
10. Hexadentate
11. E.D.T.A.

Question 3.
Match the following :

Answer:

1. (h)
2. (g)
3. (f)
4. (e)
5. (b)
6. (d)
7. (a)
8. (c)
9. (i)

Question 4.
Answer in one word/sentence :

1. Which type of isomerism is found in [CO(NH₃)₅Br]SO₄ and [CO(NH₃)₅SO₄]Br?
2. Name the organometallic compound which is used as an antiknock compound in petrol. (MP2018)
3. Complexes of E.D.T.A. formed with calcium is used to remove the poisoning caused due to which metal?
4. How is the structure of dibenzene?
5. What type of hybridization is found in [Ni(CO)₄]?
6. Which type of isomerism is represented by [Cr(H₂O)₅SCN]²⁺ and [Cr(H₂O)₅NCS]²⁺?

Answer:

1. Ionization isomerism
2. Tetraethyl lead
3. Lead
4. Sandwich
5. sp³
6. Linkage isomerism.

Coordination Compounds Very Short Answer Type Questions

Question 1.
Among the following ions, whose magnetic moment value will be maximum. (NCERT)

1. [Cr(H₂O)₆]³⁺
2. [Fe(H₂O)₆]²⁺
3. [Zn(H20)6]²⁺.

Answer:
2. [Fe(H₂O)₆]²⁺.

Question 2.
Write two examples of monodentate ligands.
Answer:
NO⁺ (Nitrosonium), NH₂NH₃ (Hydrazinium).

Question 3.
What will be the geometry of [Cr(NH₃)₆]³⁺ complex ion?
Answer:
Tetrahedral.

Question 4.
Write IUPAC name of the complex : [Pt(NH₃)₄][PtCl₄].
Answer:
Tetraammine platinum (II) tetrachloridoplatinate (II).

Question 5.
What is the hybridization of Fe in the complex ion [Fe(CN)₆]^3-?
Answer:
Hybridization of Fe in the complex [Fe(CN)₆]^3- is d²sp³.

Question 6.
State the full name of EDTA.
Answer:
Ethylene diammine tetraacetate ion.

Question 7.
Which are eg orbitals?
Answer:
d_{x² – y²} and d_z² are e_g orbitals.

Question 8.
Give an example of a neutral bidentate ligand.
Answer:
Ethylene diamine

Question 9.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligand co – ordinated with the central metal ion?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the some with respect to each other.

Question 10.
State the magnetic property and hybridization of [Ni(CO)₄J.
Answer:
Diamagnetic (All electrons paired) and sp³ hybridization.

Coordination Compounds Short Answer Type Questions

Question 1.
Explain double salt and complex salt. Give one – one example of each.
Answer:
Double salt: Double salts are additive compounds which are stable in the crystal lattice but when dissolved in water break into different compounds.
Example : Ferrous ammonium sulphate is a double salt which ionise in water as
FeSO₄ (NH₄)₂ SO₄ . H₂O ⇌ Fe²⁺ + SO₄^2- + 2(NH₄)⁺ + SO₄^2- + H₂O

Complex salt:
The compounds in which ligand with lone-pair electron are linked with any metal atom or metal ion by coordinate bonds, are called coordination compounds. In these compounds metal ion and ligand in combined state act as complex ion and thus these compounds are also known as complex compounds.
Example: K₄[Fe(CN)₆].

Question 2.
What do you understand by ligand ? Explain giving example.
Answer:
Ligand:
Any atom, ion or molecule which can donate electron pair to central ion and forms co – ordinate bond are called ligand.
Example : In K₄[Fe(CN)₆], CN is ligand. (MPBoardSolutions.com) In ligand the specific atom which donates the electron pair is known as donor atom.
On the basis of number of donor atoms in a ligand, they are classified as monodentate, bidentate, tridentate, polydentate ligands. Some such ligands are given below :

In the ligand the asterisk atom is donor atom.

Question 3.
What is meant by the chelate effect? Give an example. (NCERT)
Answer:
When a ligand attaches to the metal ion in a manner that forms a ring, then the metal – ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect for example :

Question 4.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain. (NCERT)
Answer:
H₂O is a weak ligand [Ni(H₂O)₆]²⁺ is a outer orbital complex. The complex has two unpaired electron. The d – d transition is possible. It absorbs red light and complementary green light is emitted.

CN^–is a strong ligand. The unpaired electrons are paired up. The central atoms undergoes dsp² hybridization. Square planar complex is formed. No unpaired electrons are present i.e. d – d transition is not possible, hence the complex is colourless.

Question 5.
[Fe(CN)₆]⁴⁺ and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, why? (NCERT)
Answer:
In both the complexes, Fe is in +2 state with the configuration 3d⁶ i.e., it has four unpaired electrons. As the ligand H₂O and CN^– possess different crystal field splitting energy (Δ₀), they absorb different components of the visible light (VIBGYOR) for d – d transition. Hence, the transmitted colours are different.

Question 6.
Discuss the nature of bonding in metal carbonyls. (NCERT)
Answer:
The metal carbon bonds in metal carbonyls have both s nad p characters. M – C σ – bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. M – C π – bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti – bonding π * orbital of carbon monoxide. (MPBoardSolutions.com)
This is also known as back bonding of the carbonyl group. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal. This synergic effect strengthens the bond between CO and the metal.

Question 7.
Give evidence that [CO(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionization isomers. (NCERT)
Answer:
Ionisation isomers when dissolved in water furnish different ions which can be tested.

White precipitate of AgCl indicates that the isomer has Cl ion outside the co – ordination sphere.

White precipitate of BaSO₄ indicates the isomer has SO₄^2- ion outside the co – ordination sphere.

Question 8.
Explain optical isomerism in co – ordination compounds.
Answer:
Optical isomerism:
This type of isomerism is observed in such similar compounds which are the mirror images of each other and are non- superimposable. They rotate the path of plane polarized light to the left (l) or to the right (d).

Question 9.
What is complex ion?
Answer:
Complex Ion:
A complex ion is an electrically charged redical or a species, carrying either positive or negative charge, in which the central metal ion is surrounded through co – ordinate bond by a suitable number of ligands (neutral molecules or negative ions).
Example:
Complex ferrocyanide ion[Fe(CN)₆]^4- is formed by the union of six CN^–ions with one Fe²⁺ ion. While writing the formula of a complex ion, the co – ordinating groups are written inside the bracket ( ), and the whole of the complex ion in a square bracket [ ]. The net charges is written on right hand top comer of the square bracket. The square bracket is known as co – ordination sphere.

Question 10.
Give example of bidentate and hexadentate ligand.
Answer:
Bidentate ligand :

Hexadentate ligand:

Question 11.
What is co – ordination number? Give any two examples.
Answer:
Co – ordination number:
Number of ligands which are directly linked with central metal or metal ion by co – ordinate bonds, are co – ordination number.
Example:

• [CO(NH₃)₆]Cl₃ co – ordination number of CO³⁺ ion is 6.
• In [Ag(CN)₂] co – ordination number of Ag is 2.

Question 12.
What are organometallic compounds? Write its two applications
Answer:
Those compounds in which the carbon atom of organic groups are directly bonded to metal atoms are called organometallic compounds. The compound of element such as boron, phosphorus, silicon, germanium and antimony with organic groups are also included in the organometallics.

Applications of organometallic compounds :

1. Tetraethyl lead (C₂H₅)₄Pb is used as an antiknock compound.
2. Ziegler – Natta catalysis is used in the polymerization of ethylene or other alkenes.
3. Ethyl mercuric chloride (C₂H₅HgCl) is used in agriculture as an insecticide.
4. Willkinson catalyst is used in the hydrogenation of some alkenes.

Question 13.
Explain geometrical isomerism with an example.
Answer:
Geometrical isomerism:
When the ligands are situated at different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called cis – isomer. The isomer in which the similar group occupy the opposite position (making an angle of 180° with metal ion) is called transisomer. This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) type of complexes.

Question 14.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? (NCERT)
Answer:
[NiCl₄]^2- has 2 unpaired electrons and is paramagnetic (See Q. No. 4 for details). Like CN^–, CO is also strong field ligand, same like CO, it causes pairing of electrons. No unpaired electrons left, hence it is diamagnetic.

Question 15.
[Fe(H₄O)₆ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain. (NCERT)
Answer:
In both the complexes Fe is in +3 oxidation state with the configuration 3d⁵. CN^– is a strong ligand. In its presence, 3d – electrons pair up leaving only one unpaired electron. The hybridization is d²sp³ forming inner orbital complex. H₄O is a weak ligand. In its presence, 3d – electrons do not pair up. The hybridization is sp³ d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 16.
Explain [CO(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex. (NCERT)
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state with the configuration 3d⁶. In the presence of NH₃, 3d – electrons pair up leaving two d – orbitals empty. Hence, the hybridization is d²sp³ forming an inner orbital complex.

In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state with the configuration 3d⁶. In presence of NH₃, the 3d – electrons do not pair up. The hybridization involved is sp³d² forming outer orbital complex.

Question 17.
Write the IUPAC names of the following co – ordination compounds : (NCERT)

1. [CO(NH₃)₆]C₁₃
2. [CO(NH3)5CI]C12
3. K₃[Fe(CN)₆]
4. K₃[Fe(C₂O₄)₃]
5. K₂[PdCl₄]
6. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl.

Answer:

1. Hexaammine cobalt(III) chloride
2. Pentaamminechloridocobalt(III) chloride
3. Potassium hexacyanoferrate(III)
4. Potassium trioxalatoferrate(III)
5. Potassium tetrachloridopalladate(II)
6. Diamminechlorido(methanamine)platinum(II) chloride.

Question 18.
Explain the structure of [Ni(CO)₄] on the basis of valence bond theory.
Answer:
Complexes in which the metal ion is sp³ hybridized represent tetrahedral geometry.
For example : Formation of tetracarbonylnickel (0) Can be explained by sp³ hybridization.
In [Ni(CO)₄], Nickel is in zero oxidation state. Thus, outer electronic configuration of nickel is 3d⁸ 4s³. CO is a strong ligand, thus due to the effect of ligand the 4s electrons move to 3d and make all the 3d electrons paired. This way, one 4s and three 4p become vacant and intermix to form equivalent

sp³ hybridization of [Ni(CO)₄]

sp³ hybrid orbitals which are tetrahedrally oriented. These four hybrid orbitals overlap with lone electron pairs of
4CO and form sigma bond.

[Ni(CO)₄] does not contain unpaired electron. Thus, it is diamagnetic and tetrahedral.

Question 19.
Explain the structure of [Zn(NH₃)₄]²⁺ on the basis of valence bond theory.
Answer:
Structure of [Zn(NH₃)₄]²⁺ Tetraammine zinc (II) ion:
Outer electronic configuration of zinc (Z = 30) in ground state is 3d¹⁰4s². In this complex zinc is in +2 oxidation state with the outer electronic configuration of 3d¹⁰.
The 3d orbital being completely filled does not take part in hybridization. The vacant 4s and 4p orbitals hybridize and form four hybridized sp³orbitals directed

towards the four comers of a tetrahedron. The four lone pairs from four NH₃ overlap with these sp³ orbitals and forms four σ – bonds. The compound is diamagnetic as it does not contain unpaired electrons.

Question 20.
Write IUPAC name of the following :

1. [Pt(NH₃)₃ Cl₂]
2. K₃[Fe(CN)₆]
3. [CO(NH₃)₆] Cl₃
4. Pt[(NH₃)₆] Cl₄
5. CuCl₃.

Answer:

1. Dichlorodiammine platinum (II)
2. Potassium hexacyanoferrate (III)
3. Hexaammine cobalt (III) chloride
4. Hexaammine platinum (IV) chloride
5. Tetrachlorocuprate (II)

Question 21.
1. Write I.U.P.A.C. name of

2. Write the ligands and co – ordination number of the complex [Cr (NH₃)₄(ONO) CI]NO₃.
3. Write chemical formula of carbonate pentaammine cobalt (III) chloride.
Answer:
1. µ – amido – µ -hydroxobis – (tetraammine cobalt) (III) ion.
2. Ligands :

• NH₃
• ONO
• Cl

Thus, number of ligands are 3 and co – ordination number is 6.

3. [CO(NH₃)₅CO₃]Cl.

Question 22.
Using IUPAC norms write the formulas for the following :

1. Tetrahydroxozincate(II)
2. Potassium tetrachloridopalladate(II)
3. Diamminedichloridoplatinum(II)
4. Potassium tetracyanonickelate(II)
5. Pentaamminenitrito-o-cobalt(III)
6. Hexaamminecobalt(III) sulphate
7. Potassium tri-(oxalato)chromate(III)
8. HexaamminepIatinum(IV)
9. Tetrabromidocuprate(II)
10. Pentaamminenitrito-N-cobalt(III).

Answer:

1. [Zn(OH]²
2. K₂[PdCl4]
3. [Pt(NH₃)₃Cl₂]
4. K₂[Ni(CN)₄]
5. [CO(NH₃)₅(ONO)]²⁺
6. [CO(NH₃)₆]₂ (SO₄)₃
7. K₃[Cr(C₂O₄)₃]
8. [Pt(NH₃)₆]⁴⁺
9. [Cu(Br)₄]^2-
10. [CO(NH₃)₅(NO₂)]²⁺.

Question 23.
Write the I.U.P.A.C. names of the following compounds:
(i)
(a) [HgI₄]^2-
(b) [Ag (CN)₂]^–
(c) [Fe (C₅H₅)₂]
(d) K [Ag (CN)₂].

(ii) What is Zeise’s Salt and Ferrocene? Explain with structure.
Answer:
(i)
(a) Tetraiodomercurate (II) ion.
(b) Dicyanoargentate (I) ion.
(c) Bis (cyclopentadienyl) iron (II).
(d) Potassium dicyano argentate (I).

(ii)
(a) Zeise’s salt K [PtCl₃ – η²(C₂ H₄ ) :
This salt was prepared by Danish pharmacist Zeise in 1830. It is one of the compound of transition metals which was prepared earlier. The plane of ethylene molecule and the C = C axis are perpendicular to the expected bond direction of the central atom.

(b) Ferrocene Fe (η⁵ – C₅H₅)₂:
It is an orange yellow coloured compound. Kealy and Pauson reported it in 1951. It has sandwich structure in which iron atom is in between two cyclopentadienyl rings. The planes of the rings are parallel so that all the carbon atoms are equidistant from iron atom.

Question 24.
Explain Linkage and Ionisation isomerism with example.
Answer:
Linkage isomerism:
Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligand.

In structure I Ni²⁺ is linked by thiocyanate sulphur and in structure II it is linked by nitrogen atom.

Ionization Isomerism:
This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
[CO(NH₃)₅Br] SO₄ – It liberates SO^2- ions.
[CO(NH₃)₅SO₄]Br – It liberates Br^– ions.

Question 25.
What is the difference between primary and secwraary valency? Give example also.
Answer:
Primary valency is ionisable while secondary valency does not ionise. Primary valency is represented in figure by solid lines and secondary valency by broken or dotted lines.

Example:
[CO (NH₃)₆] Cl₃, primary valency is 3 and secondary valency is 6. Co – Central metal, NH₃, Cl – Ligand.

Question 26.
What is effective atomic number (EAN)?
Answer:
In co – ordination compounds, sum of electrons present in central metal ion and electrons received in bond formation is called effective atomic number.
EAN = Atomic number – Electrons lost in ion formation + Electrons obtained in bond formation
In K₄[Fe(CN)₆] EAN for Fe = 26 – 2 + 12 = 36.

Coordination Compounds Long Answer Type Questions

Question 1.
Explain the bonding in co – ordination compounds in terms of Werner’s postulates. (NCERT)
Answer:
Werner’s co – ordination theory:
Alfred Werner gave his co – ordination theory in 1893. The important postulates of this theory are :
1. All metals in atomic or ionic form exhibit two types of valencies in co – ordination compounds :

• Primary or principal or ionic valency ( ….. )
• Secondary or auxiliary or non-ionic valency (-).

The primary valency is ionizable and it is shown by dotted lines. The secondary valency is non – ionizable and it is shown by continuous line.

2. Primary valency represents oxidation states of metal atom or ion and secondary valency represents co – ordination number of metal ion which is fixed for a particular atom.

3. The primary valencies are satisfied by negative ions whereas the secondary valencies may be satisfied either by negative ions (Example Cl^–, Br^–, CN^– etc.) or neutral molecules (Example H₂O).

4. Secondary valencies are directed towards fixed position in space.

5. Every element tends to satisfy both its primary and secondary valencies. For this
purpose a negative ion may often act a dual behaviour i.e., it may satisfy primary as well as secondary valency ( ).

Example:
Luteo cobaltic chloride COCl₃.6NH₃ or [CO(NH₃)₆]Cl₃.
Purpureo cobaltic chlorideq COCl₃.5NH₃ or [CO(NH₃)₅Cl]Cl₂

Question 2.
List various types of isomerism possible for co – ordination compounds, giving an example of each. (NCERT)
Answer:
Isomerism in Co – ordination Compounds:
Two or more compounds having the same molecular formula but different arrangement of atoms are called isomers.
Isomerism in Co – ordination Compounds is given below:

1. Structural Isomerism
This type of isomerism arises due to the difference in structures of co-ordination compounds. It is further sub divided into four different types as explained below :

(a) Ionization Isomerism:
This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
[CO(NH₃)₅Br] SO₄ – It liberates SO^2- ions.
[CO(NH₃)₅SO₄]Br – It liberates Br^– ions.

(b) Co – ordination isomerism:
This type of isomerism is shown by such compound which contain both cationic and anionic species and it arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
For example, [CO(NH₃)₆][Cr(CN)₆ and [Cr(NH₃)₆][CO(CN)₆]
hexaamminecobalt (III) hexacyano chromate (III) and hexaamminechromium (III) hexacyanocobalt (III).

(c) Linkage isomerism:
Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligand.

(d) Hydrate Isomerism:
This isomerism arises when different number of water molecules are present within and outside the co-ordination sphere. For example, three hydration isomer of CrCl₃.6H₂O are :

(i) [Cr(H₂O)₆]Cl₃ All the six water molecules act as ligand
(Violet)
(ii) [Cr(H₂O)₅Cl]Cl₂.H₂O Five water molecules acts as ligand while one molecule
(Green)
of water as crystallization.
(iii) [Cr(H₂O)₄Cl₂]Cl.2H₂O Four water molecules acts as ligand, while the two molecules of water as crystallization.
(Green)

2. Stereoisomerism:
Two compounds are called stereoisomers when they contain the same ligand in their co – ordination sphere but differ in their spatial arrangement. Stereoisomerism is further classified as geometrical and optical isomerism.

(a) Geometrical Isomerism:
When the ligands are situated at different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called cis – isomer. (MPBoardSolutions.com) The isomer in which the similar group occupy the opposite position (making an angle of 180° with metal ion) is called transisomer. This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) type of complexes.

(b) Optical Isomerism:
This type of isomerism is observed in such similar compounds which are the mirror images of each other and are non- superimposable. They rotate the path of plane polarized light to the left (l) or to the right (d).

Question 3.
Explain crystal field theory.
Answer:
This theory was proposed by Bethe and Von Black. According to this theory bonding between central metal ion and ligand is due to pure electrostatic attraction. If ligand is anion then attraction towards cation, is just like the attraction between two oppositely charged particles. If ligand is neutral molecule then the anionic end of this dipole is attracted to central positive ion. Thus, bonding between them is due to ion – ion attraction or ion dipole attraction.

In crystal field theory due to approaching ligands the d – orbitals split into different energy levels on the basis of extent of splitting (depend on metal ion and nature of ligand) structure and properties of complex can be discussed. (MPBoardSolutions.com) Colour of transition metal complex is due to absorption of visible light which leads to excitation of electron from one d – orbital to the other d – orbital (d – d transition). This way this theory is easy and successfully explains maximum properties of the complexes.

MP Board Class 12th Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Aniline is generally purified by:
(a) Steam distillation
(b) Simple distillation
(c) Distillation under reduced presence
(d) Sublimation
Answer:
(a) Steam distillation

Question 2.
Glycerol boils at 290°C with slight decomposition. Impure glycerol is purified by:
(a) Steam distillation
(b) Simple distillation
(c) Vacuum distillation
(d) Extraction with solvent
Answer:
(c) Vacuum distillation

Question 3.
The blue or green colour obtained in Lassaigne’s test is due to the formation of:
(a) NaCN
(b) Na₄[Fe(CN)₆]₄
(c) Fe₃[Fe(CN)₆]₄
(d) Fe₄[Fe(CN)₀]₃
Answer:
(d) Fe₄[Fe(CN)₀]₃

Question 4.
In qualitative analysis of organic compound by Lassaigne’s test the violet colour obtained with sodium nitroprusside indicate the presence of:
(a) Nitrogen
(b) Sulphur
(c) Oxygen
(d) Halogen.
Answer:
(b) Sulphur

Question 5.
The blood red colour compound formed during the qualitative analysis of nitrogen and sulphur together is:
(a) Fe₄[Fe(CN)₆]₂
(b) Fe(SCN)₃
(c) KSCN
(d) Na₂S.NaCN.
Answer:
(b) Fe(SCN)₃

Question 6.
Kjeldahl’s as method is used for estimation of:
(a) Sulphur
(b) Netrogen
(c) Halogen
(d) Oxygen
Answer:
(b) Netrogen

Question 7.
A compound with empirical formula C₂H₅O had molecular mass 90. The formula of compound is:
(a) C₄H₁₀O₂
(b) C₂H₅O
(c) C₃H₆O₃
(d) C₅H₁₄O
Answer:
(a) C₄H₁₀O₂

Question 8.
The amount of sulphur present in an organic compound is estimated by changing into:
(a) H₂S
(b) SO₂
(c) H₂SO₄
(d) H₂SO₄
Answer:
(d) H₂SO₄

Question 9.
The reagent used in Carius method to estimate halogen is:
(a) HNO₃ and HCl
(b) HNO₃ and H₂SO₄
(c) Fuming HNO₃ and BaCl₂
(d) Fuming HNO₃ and AgNO₃
Answer:
(d) Fuming HNO₃ and AgNO₃

Question 10.
The gas collected in Duma’s method to estimate of nitrogen in organic compound is:
(a) N₂
(b) NO
(c) NH₃
(d) None of these
Answer:
(a) N₂

Question 11.
An organic compound contain C = 80% and H = 20%. The compound shall be:
(a) C₆H₆
(b) C₂H₅ – OH
(c) C₂H₆
(d) CHCl₃
Answer:
(c) C₂H₆

Question 12.
An organic compound contain C = 39-9%, H = 6‘7% and O = 53.4%. The graphical formula shall be:
(a) CHO
(b) CHO₂
(c) CH₂O₂
(d) CH₂O
Answer:
(d) CH₂O

Question 13.
In an organic compound the ratio of mass is C:H:O = 4:1:5. Its empirical formula shall be:
(a) C₂HO
(b) C₂H₄O₄
(c) CH₄O₂
(d) CH₃O
Answer:
(d) CH₃O

Question 14.
The main source of organic compound is:
(a) Coaltar
(b) Petroleum
(c) Both
(d) None of these
Answer:
(c) Both

Question 15.
But – 1,2 diene contains:
(a) Only sp – hybridized carbon atom
(b) Only sp² – hybridized carbon atom
(c) sp and sp² hybridized carbon atom
(d) sp, sp² and sp³ hybridized carbon atom
Answer:
(d) sp, sp² and sp³ hybridized carbon atom

Question 16.
I.U.P.A.C. name of

(a) 2 – Ethyibut – 2ene
(b) 3 – Ethylbut – 2 – ane
(c) 2 – Methylpent – 3 – ene
(d) 3 – Methylpent – 2 – ene
Answer:
(d) 3 – Methylpent – 2 – ene

Question 17.
The I.U.P.A.C. name of the compound having the formula of Cl₃ – CCH₂CHO is:
(a) 3,3,3 – trichloropropan – l – al
(b) 1,1,1 – trichloropropan – l – al
(c) 2,2,2 – trichloropropan – 1 – al
(d) Chloral.
Answer:
(a) 3,3,3 – trichloropropan – l – al

Question 2.
Fill in the blanks:

1. Paper chromatography is based on the law of …………………………….
2. Column chromatography is based on the law of …………………………….
3. Aniline is purified by ………………………… method.
4. Hybridisation of central carbon atom of a carbocation is ……………………………..
5. Benzoic acid is purified by …………………………….
6. Before test of halogen, sodium extract is heated with ……………………………
7. Isomerism found in organic compounds of same series is ………………………………..
8. Chemical name of freon organic compound which is used in air conditions and refrigerators is ………………………….. and its chemical formula is ……………………..
9. ……………………….. ratio of elements in a compound is called its empirical formula.
10. The process of fractional crystallization of separation of two substances depending on the difference of ……………………………….
11. In Lassaigne’s test, blue or green colour is due to the formation of ……………………………….
12. On adding FeCl₃ solution to sodium extract ……………………………. colour is obtained. The name of the compound is …………………………..
13. In organic compound, presence of amount of halogen can be detected by converting it into …………………………………….
14. A compound contain 80% carbon and 20% hydrogen, its formula will be ……………………………………
15. …………………………. gas is produced by the action of water on calcium carbide.
16. R – CONH₂ is an ……………………………
17. Marsh gas mainly contain …………………………….. gas.

Answer:

1. Distribution
2. Adsorption
3. Steam distillation
4. Sp²
5. Sublimation
6. Cone. HNO₃,
7. Metamerism
8. Difluorodichloro methane CF₂Cl₂
9. Simplest
10. Solvent
11. Ferri – ferro cyanide
12. Red, ferric sulphocyanite
13. Silver halide
14. C₂H₆
15. Acetylene
16. Amide
17. Methane.

Question 3.
Answer in one word/sentence:

1. Method used for separation of components on the basis of adsorption is known as?
2. What is conversion of solid substance on heating into vapours without changing into liquid known as?
3. Which element is detected by Duma’s method?
4. What is method of obtaining pure substance by vaporisation of impure liquid followed by condensation of vapours known as?
5. What is the charge on carbon in carbanion?
6. Which formula represents ratio of atoms of elements present in a molecule of a substance?
7. What is the nature of nucleophile?
8. What are cations carrying positive charge on carbon known as?
9. What is the nature of electrophile?
10. Which effect is responsible for the displacement of electrons of covalent bond towards or aways from carbon atom in an organic molecule?
11. Mixture of KMnO₄ and KOH is known as?

Answer:

1. Chromatography
2. Sublimation
3. Nitrogen
4. Distillation
5. Negative
6. Empirical formula
7. Electronegative
8. Carbocation
9. Electropositive
10. Inductive effect
11. Baeyer’s reagent.

Question 4.
Match the following:
[I]

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

[II]

Answer:

1. (b)
2. (d)
3. (e)
4. (a)
5. (c)

[III]

Answer:

1. (a)
2. (d)
3. (b)
4. (c)

Organic Chemistry: Some Basic Principles and Techniques Very Short Answer Type Questions

Question 1.
What is the chemical name of CH₃ – CH₂ – CHCl – CH₃?
Answer:
iso – butyl chloride.

Question 2.
What is the mixture of KMnO₄ and KOH called?
Answer:
Bayer’s reagent.

Question 3.
IUPAC name of Vinegar?
Answer:
Ethanoic acid.

Question 4.
What is the IUPAC name of grain alcohol?
Answer:
Ethanol.

Question 5.
For the mixture of CuSO₄ and Camphor, Camphor is separated by?
Answer:
Sublimation.

Question 6.
How purification of naphthalene was done?
Answer:
By sublimation.

Question 7.
The purification by Column chromatography occur because?
Answer:
Different absorption.

Question 8.
Purification of petroleum?
Answer:
Fractional distillation.

Question 9.
Balsentein test performed for?
Answer:
In halogen detection.

Question 10.
Free radicals are formed by?
Answer:
Homolytic fission.

Question 11.
Main source of organic compounds are?
Answer:
Coaltar and petroleum.

Question 12.
General formula of alcohol is?
Answer:
C_nH_2n+1OH.

Question 13.
What is Chiral molecule?
Answer:
Those which are not superimposable on their mirror images.

Question 14.
What is the name of the compound Cl – CH₂ – CH₂ – COOH?
Answer:
3 – Chloro propanoic acid.

Question 15.
Write the structural formula of iso – butyl chloride?
Answer:
CH₃CH₂CHClCH₃.

Question 16.
C_nH_2n-2 is formula of?
Answer:
Alkynes.

Question 17.
What is Bayer’s reagent?
Answer:
Alkaline KMnO₄.

Question 18.
Compounds different in configuration are called?
Answer:
Stereo isomers.

Question 19.
What is the IUPAC name of Cl₃C.CH₂CHO?
Answer:
3,3,3 – trichloro propanol.

Question 20.
Which type of isomerism is found in nitro ethane?
Answer:
Tautomerism.

Question 21.
Write the name and formula of Freon?
Answer:
Difluoro – dichloro methane (CF₂Cl₂).

Question 22.
The mixture of o – nitrophenol and p – nitrophenol is separated by which method?
Answer:
Vapour distillation method.

Question 23.
Which gas is present in Marsh gas?
Answer:
Methane.

Question 24.
The decomposition of glycerine occurs before its b.p., by which method it can be purified?
Answer:
Low pressure distillation.

Question 25.
Formalin is formed by which compound?
Answer:
HCHO.

Question 26.
What is the structural formula of gem – dihalide?
Answer:

Question 27.
What is the mechanism of this reaction:
CH₃CH₂I + KOH_(aq) → CH₃CH₂OH + KI.
Answer:
Nucleophilic substitution.

Question 28.
Kjeldahl’s method is used for estimation of which element?
Answer:
Nitrogen.

Question 29.
What is Elution ?
Answer:
The process of separation of products by different absorption rate is called elution

Question 30.
Which is the latest and better technique for the separation and purification of organic compounds?
Answer:
Chromatography method.

Organic Chemistry: Some Basic Principles and Techniques Short Answer Type Questions

Question 1.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
For testing sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. If H₂SO₄ were used, lead acetate itself will react with H₂SO₄ to form white ppt. of lead sulphate which interfere the test.

Question 2.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
Carbon dioxide is acidic and it reacts with strong base KOH to form potassium carbonate
2KOH + CO₂ K₂CO₃ + H₂O

This results in increase in mass of potassium hydroxide from the increase in mass of CO₂ produced, the amount of carbon in the organic compound can be calculated by using the formula:

Question 3.
What is homologous series? Write its characteristics?
Answer:
Homologous series is a series of similarly constituted organic compounds in which the members possess the same functional group, have similar or almost similar chemical characteristics, can be represented by the same general formula and the two consecutive members differ by CH₂ group in their molecular formulae.

The various members of a particular homologous series are called homologues. A few homologues of alcohol series (containing straight chain alcohols) are as follows: General formula C_nH_2n+1OH

Characteristics of homologous series:

1. All the members of a series can be represented by the general formula. For example, general formula of alcohol family is C_nH_2n+1OH.
2. The two successive members of a particular family is differ by – CH₂ group or by 14 atomic mass unit (12 + 2 × 1).
3. Different members in a family have common functional group, for example, alcohol family given above.
4. The members of a particular family have almost identical chemical properties and their physical properties such as melting point, boiling point, density, solubility etc. show a proper gradation with the increase in the molecular mass.
5. The members present in a particular series can be prepared almost by similar methods known as the general methods of preparation.

Question 4.
What are primary, secondary, tertiary and quarternary C of organic compound?
Answer:
Primary carbon atom:
Carbon atom in the organic compound which is linked with only one carbon atom is called primaty (p) or (1°) carbon atom.

Secondary carbon atom:
Secondary carbon atom is that carbon atom which is linked with two more carbon atoms in the compound. It is also represented by (2°) or (s) carbon atom.

Tertiary carbon atom:
The carbon atom which is linked with three more carbon atoms, is called tertiary (3°) or (t) carbon atom.

Quarternary carbon atom:
The carbon atom which is linked with four other carbon atoms, is called quartemary carbon atom. It is denoted by 4° or q.
In the following example primary (p), secondary (s) and tertiary (t) carbon atoms are represented:

Where, p = primary, s = secondary, t = tertiary and q = quartemary.

Question 5.
What is resonance? Write its applications?
Answer:
Sometimes it is found that all the known properties of a compound cannot be explained by one structure and for such compounds we draw two or more structures. (MPBoardSolutions.com) Such structures are called resonating structures or canonical forms or contributing structures and the phenomenon is called resonance or mesomeric effect. This is a permanent effect. This effect is transmitted through the chain. There are two types of resonance or mesomeric effect:

1. + R or + M effect:
A group is said to have +R or +M effect when the displacement of the electron pair is away from it. For example,

2. – R or – M effect:
A group is said to have – R or – M effect when the displacement of the electron patr is towards it. For example,

Uses:

1. For determination of structure of benzene.
2. To explain dipole moment.
3. To explain the strength of acids and bases.

Question 6.
What is the reason, that carbon forms large number of compounds?
Or, Explain the special properties of carbon,
Answer:
Anomalous behaviour of First Element of the Group (Carbon):
Carbon the first member of group – 14 shows an anomalous behaviour i.e., differ from the rest of the members of its family. The main reasons for this difference are:

1. Small atomic and ionic size
2. Higher electronegativity
3. Higher ionisation enthalpy
4. Absence of d – orbital in the valence shell

The main points of difference are:

1. It is an important component of animal kingdom.

2. It is also found in free state in nature.
3. It possesses the property of catenation because C – C bond energy is very high 353.3 kJmol^-1.

4. Carbon exist in various allotropic form. Its three crystalline form are diamond, graphite and fullerene.

5. Carbon atom has tendency to form pπ – pπ bond with other carbon atom and also with oxygen, nitrogen, sulphur etc. Due to this, carbon – carbon, carbon – oxygen, carbon – nitrogen, etc. double and triple bonds are possible.

6. Carbon is the only element which forms highly stable open chain, cyclic hydrocarbon and aromatic hydrocarbon with hydrogen.

It is due to its property called catenation. It is the ability of like atoms to link with one another through covalent bonds. This is due to smaller size and higher electronegativity of carbon atom and unique strength of carbon – carbon bond. (MPBoardSolutions.com) Since the bond energy of C – C bond is very large (348 kJ mor1). Carbon forms long straight or branched C – C chains or rings of different size and shape. However, as we move down the group the element – element bond energies decreases rapidly viz C – C (348 kJ mol^-1), Si – Si (297 kJ mol^-1), Ge – Ge (260 kJ mol^-1), Sn – Sn (240 kJ mol^-1), Pb – Pb (81 kJ mol^-1), and therefore, the tendency for catenation decreases in the order:
C >> Si > Ge = Sn > Pb.

7. Carbon forms three types of oxide, monoxide, dioxide and suboxide. Bond energy of carbon monoxide is highest among diatomic molecules.

8. Carbon atom can link with other metals directly through covalent bonds compound formed are called organometallic compound.

Question 7.
Explain metamerism and tautomerism with example?
Answer:
Metamerism:
The compounds having same molecular formula but different number of carbon atoms (or alkyl group) on either side of the functional group are called metamers and phenomenon is called metamerism.

Tautomerism:
This is a special type of functional isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other. For example, acetaldehyde and vinyl alcohol are tautomers.

Question 8.
What are Nucleophile? Explain with example?
Answer:
Nucleophiles:
The species having an atom, unshared or ione pair of electron and seeking positive sets are called nucleophiles.
Neutral nucleophiles: NH₃, H₂O, R – O – R.
Negative nucleophiles: Cl^–, OH^–, NH₂^–, CN^–.

Question 9.
What are Electrophiles? Explain with example?
Answer:
Electrophiles:
The positively charged or neutral species which are deficient of electron and can accept, lone pair of electron are called electrophiles.

Neutral electrophiles:
BF₃, AlCl₃, FeCl₃.

Negative electrophiles:
H₃O⁺, Cl⁺, NO₂⁺.

Question 10.
How nitrogen is tested in any organic compound by Lassaingen’s method?
Answer:
Take 2 ml of Sodium extract, add a 2 ml of freshly prepared solution of ferrous sulphate along with 1 – 2ml of NaOH. Heat and then cool the solution. (MPBoardSolutions.com) Green precipitate of Fe(OH)₃ is obtained. Add cone. HCl so that green precipitate of ferrous sulphate goes into solution. Then 2 – 3 drops of ferric chloride solution is added. If green or blue colour is obtained then the substance contain nitrogen.

Question 11.
How will you test the presence of sulphur in any organic compounds?
Answer:
Sulphur test:
1. Sodium nitropruside solution is added to the sodium extract if violet colour appears. Confirms the presence of sulphur in given organic compound.

2. Sodium extract is acidified with acetic acid and lead acetate solution is added. If black precipitate is obtained confirms the presence of sulphur in organic compound.

Question 12.
A sample of 0.50g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in SO ml of 0.5M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralization.Find the percentage composition of nitrogen in the compound?
Solution:
Volume of acid taken = 50 ml of 0.5 M H₂SO₄
= 25 ml of 1.0 M H₂SO₄
Volume of base used for neutralization of acid
= 60 ml 0.5 M NaOH
= 30 ml 0.1 M NaOH

∴ 30 ml of 1.0 M NaOH = 15 ml of 1.0 M H₂SO₄
∴Volume of acid used by ammonia = 25 – 15 = 10 ml
% Amount of Nitogen = image 17
% N = \(\frac{1.4 × 2 × 10}{0.5}\) = 5 gmN.

Question 13.
Steam distillation is useful for which organic compounds? Explain with example?
Answer:
Steam distillation:
Steam distillation is used to purify those organic compounds which are practically immiscible with water, volatile in steam and has fairly high vapour pressure (low boiling point). In this method, the impure liquid is taken in a heated flask and steam is passed over it with the help of a steam generator (Fig.). (MPBoardSolutions.com) The mixture of steam and the volatile organic compound is condensed and collected in a receiver. From this mixture, water is removed by using separating funnel.

During steam distillation, a liquid boils only when the sum of vapour pressure of the liquid (P₁) and of water (P₂) becomes equal to the atmospheric pressure than its boiling point e.g., a mixture of water and a steam volatile insoluble substance will vaporise close below 373K. (MPBoardSolutions.com) This above technique is used for separating aniline from aniline water mixture and also for separation of p – nitro phenol from p – nitro phenol (o – nitrophenol is steam volatile).

Question 14.
What is the principle of Adsorption chromatography? Explain?
Answer:
Chromatography:
The process by which different components of a mixture are separated by distributing in stationary or mobile phases on the basis of difference in adsorption abilities on any adsorbent, is called chromatography.

Adsorption chromatography:
It is also known as column chromatography. It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

Question 15.
0.2 gm of chlorine containing organic substance gave in Carius method 0.2870 gm of AgCl. Determine the percentage of chlorine in the compound?
Weight of organic substance = 0.2 gm.
Weight of AgCl = 0.2870 gm.
Percentage of Chlorine = \(\frac{35.5}{143.5}\) ×
= \(\frac{35.5}{143.5}\) × \(\frac{0.2870 × 100}{0.2}\)
= 35.5%

Question 16.
What is principle of vaccum distillation or distillation under reduced pressure 7 Explain with figure?
Answer:
Distillation under reduced pressure:
Many substances decompose at their boiling points. Hence, they cannot be purified by simple distillation. These compounds are distilled at low temperatures and low pressures. This is known as reduced pressure distillation. (MPBoardSolutions.com) Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure. This means that by lowering the pressure to which a liquid is subjected, the boiling points of the liquid can be lowered. Similarly, if the pressure is increased, the boiling point also increases. This means that a liquid can be made to boil any temperature by varying the pressure.

As shown in the figure distillation under reduced pressure is carried out in a specially designed flask called Claisen flask. (MPBoardSolutions.com) Pressure in the receiver is reduced by vacuum pump which reduces pressure in distillation flask also and liquid begins to boil at low temperature. For example, glycerol is also distilled under reduced pressure. Its b.p. is 290°C, but using 12 mm pressure it can be distilled at 180°C.

Question 17.
How halogens are detected in organic compound?
Answer:
AgNO₃ Test:
If on adding HNO and AgNO₃ in sodium extract, white ppt. comes, then AgCl is present. If the white ppt. is soluble in excess of NH4C1 than Cl is present. On adding dil. HNO₃ and AgNO₃ in sodium extract, if yellow ppt. appears than bromine and iodine is present.

Question 18.
Write the difference between Inductive effect and Electrometric effect:
Inductive effect:

1. It is a permanent effect.
2. This arises due to displacement or σ – bond.
3. Partial negative or positive charge develops.
4. Displacement reaction occurs.
5. Always present in molecule.

Electrometric effect:

1. It is temperory effect.
2. Arises due to displacement of or π – bond.
3. Complete positive and negative charge develops.
4. Addition reaction occurs.
5. This effect arises due to presence of attacking reagent.

Organic Chemistry: Some Basic Principles and Techniques Long Answer Type Questions:

Question 1.
How halogen is detected in an organic compound?
Answer:
Estimation of Halogens Carius method:
In this method estimation of halogens (Cl, Br and I) is done. A known weight of organic compound containing halogen is heated with AgNO₃ and fuming nitric acid. Carbon, hydrogen and sulphur present in the compound gets oxidized and halogens form AgX (Silver halide).

Caluculations:
Suppose = W gm
Weight of organic substance = m gm
Weight of silver halide = x gm
Molecular mass of silver halide = (108 + x) gm
Molecular mass of silver halide = x gm
∴ m gm of silver halide contain halogen = \(\frac{x}{(108 + x)}\) × m gm
W gm of substance contain halogen = \(\frac{x × m}{(108 + x)}\) gm
∴ 100 gm of organic substance contains = \(\frac{x × m × 100}{(108 + x)}\) × W
Hence, percentage of halogen

Question 2.
Write short notes on carbanion?
Answer:
Carbanion:
Carbanions may be defined as negatively charged ions, in which carbon is having negative charge and it has eight electrons in the valence shell e.g.,

Generation of carbanions:
These are mostly generated in the presence of a base by heterolytic cleavage.

Types of alkyl Carbions:
Depending on the carbon bearing negative charge carbions may be of threee types,

Orbital structure of carbanion:
The negatively charged carbon atom in a carbanion is spl hybridized. It is expected to have a tetrahedral geometry. The three hybridized orbitals with one electron each are involved in the σ – bonds with the orbitals of other atom or groups. (MPBoardSolutions.com) The fourth hybridized orbital has overlapped. It is responsible for the negative charge on carbanion and also for the distortion of its geometry. The H actual shape of the carbanion is pyramidal.

Stability of carbanion:
The stability of carbanion can be discussed with the help of inductive effect.

Question 3.
What is Inductive effect? Write its uses?
Answer:
Inductive effect:
In a covalent band between two disimilar atoms having different electro negativities the electron pair does not remain in the centre but gets attracted towards the more elecronegative atoms. (MPBoardSolutions.com) The bond becomes some what polar due to unequal sharing of the electron pair. For example, in the bond

if X is more electronegative than C, the electron pair gets attracted towards S. This shifting of electrons develops a partial negative charge denoted by on δ^– on X and C attains a partial positive charge doneted by δ⁺. Thus,

Now consider a log chain of carbon atoms with a more electronegative element say chlorine attached at one end.

The electron pair of the bond between C₁ and X gets displaced towards electronegative chlorine atom. This results in developing of partial negative charge on chlorine and partial charge on carbon. This displacement is further transmitted to other carbon atoms of the chain but the magnitude of displacement goes on decreasing with the increases in the distance of the carbon atoms from the chlorine atom as shown below:

Thus, it can be concluded that a polar bond induces polarity in the other covalent bonds in a chain. This type of displacement of electrons is referred to as inductive effect (or I effect) or transmission effect. (MPBoardSolutions.com) Thus, inductive effect may be defined as, the permanent displacement of electrons along the chain of carbon atoms due to presence of polar covalent bond in the chain.

Types of inductive effect:
There are two type of inductive effects:

1. Electron withdrawing inductive effect (- I effect):
If the substantiates attached to the end the carbon chain is electron withdrawing, the effect is called – I effect. The decreasing order or – I effect of some atoms or groups is as follows:

2. Electron releasing inductive effect (+1 effect):
If the substantiates attached to the end of the carbon chain is electron releasing, the effect is called +I effect. Alkyl groups are electron releasing in nature. Thus, the decreasing order of +I effect is as follows:

Question 4.
Write the differences between electrophilic and nucleophilic reagents?
Electrophilic Reagent:

1. Deficient electrons.
2. Generally electrons are present in valence shell.
3. They are positive ions.
4. Neutral molecules with incomplete octet accept electrons.
5. They are Lewis acids.

Nucleophilic Reagents:

1. More electron present.
2. Generally 8 electrons are present in valence shell.
3. Negatively charged ion.
4. They are electron pair donor.
5. They are Lewis base.

Question 5.
Explain the Column chromatography technique for purification of organic compounds?
Answer:
Column chromatography:
It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

Column chromatography:
There are three steps of column chromatography:

1. Preparation of adsorbent column:
A long tube like burette is filled with a paste of a suitable adsorbent like activated Solvent Separation Continue elution magnesia, alumina, gypsum, silica gel, kieselguhr, etc. in a suitable organic solvent. The paste is prepared in that solvent in which the solution of the mixture to be separated, is prepared. When adsorbent is set, solvent is allowed to flow down.

2. Process of adsorption:
The substance or the mixture to be adsorbed is dissolved in least quantity of the nonpolar solvent like petroleum ether, benzene, etc. The solution is allowed to flow down the column. The different components of mixture get adsorbed in different parts of column. (MPBoardSolutions.com) The compound which is strongly adsorbed remains in the upper part of the column forming a band. Each component of mixture forms a separate band at a definite place. The coloured band formed are seen clearly. In case the bands are not seen due to being colourless, they are made visible by using a suitable indicator.

3. Elution:
In this process the adsorbed substance is extracted by a suitable solvent. The solvent used for this purpose is called eluent and the process is called elution. The solvents are used in the order of increasing polarity. Solvent in the increasing order of polarity are petroleum ether, petroleum ether containing benzene, alcohol with ether and pure ether.

These solvents are added one after the other. The substance which has been least adsorbed gets extracted with a solvent which is least polar while the component which has been adsorbed more strongly than others, is extracted by more polar solvent like alcohol. (MPBoardSolutions.com) By this way various components of mixture can be separated on several steps.

The various components can be separated from solvent by distillation or by using separating funnel. This technique is employed for separation of complex compounds like vitamins and hormones. The method is also employed for determination of purity of substance.

Question 6.
Write the IUPAC name of following compounds:

Answer:

Question 7.
Difference betweeen Aliphatic compound and Aromatic compound?
Answer:
Aliphatic compound:

1. These are open chain compound.
2. Generally C – C bond present.
3. Highly reactive.
4. Halogenation, nitration, sulphonation easily not occur.
5. Combustion energy high.
6. – OH group is neutral.

Aromatic compound:

1. These are closed chain compound.
2. Conjugated single and double bond present.
3. Less reactive.
4. Halogenation, nitration, sulphonation occur.
5. Combustion energy low.
6. – OH group is acidic in nature.

Question 8.
Explain the Duma’s method of determination of nitrogen in organic compound?
Answer:
Duma’s method:
This method can be used for estimation of nitrogen in all types of nitrogenous compounds. A known amount of nitrogen containing organic compound is heated with cupric oxide (CuO) in an atmosphere of CO₂. C and H₂O are oxidised to CO₂ and H₂O while N₂ gas is set free.
C + 2Cuo → CO₂ + 2Cuo
H₂ + Cuo → H₂O + Cu (in organic substance)
Nitrogen + CuO → N₂ + Some amounts of some oxides of N₂
A general equation for nitrogen containing compound is given below:
C_xH_yN_z + (2x + \(\frac{y}{2}\)) CuO → xCO₂ + \(\frac{y}{2}\) H₂O + \(\frac{z}{2}\) N₂ + (2x + \(\frac{y}{2}\)) Cu
If sulphur is present in the organic compound. It is converted into S02. During the above reaction, some oxides of nitrogen also be formed. (MPBoardSolutions.com) Therefore, the gaseous mixture is passed over heated reduced copper gauze which converts oxides of nitrogen back to nitrogen.
2NO + 2Cu → 2CuO + N₂
2NO₂ + 4Cu → 4CuO + N₂
The gaseous mixture containing CO₂, H₂O, SO₂ and N₂ is collected in a graduated nitrometer containing KOH solution. Water vapours are condensed whereas CO₂ and SO₂ are absorbed by KOH solution. Nitrogen is collected in the upper part of nitrometer. Volume of nitrogen is noted at room temperature and pressure.

Apparatus:
The main part of apparatus is combustion tube. It is a long tube, open at both the ends. The tube is packed with

1. Oxidized copper gauze which prevents backward diffusion of gases produced during combustion
2. CuO containing weighed amount of organic compound
3. Coarse CuO which oxidises the organic compound into CO₂, H₂O, SO₂ etc.,
4. A reduced copper oxide i.e., copper which converts oxides of nitrogen (NO, NO₂ etc.) back to nitrogen.

Oxides of nitrogen when passed over hot reduced copper gauze change to nitrogen by losing their oxygen.
Oxides of nitrogen + Cu → N₂ + CuO

The nitrogen set free is passed through SchifFs nitrometer filled with 40% solution of caustic potash. Caustic potash solution absorbs CO₂ while water formed gets condensed.
After the completion of combustion again CO₂ is passed through combustion tube to drive out all the remaining nitrogen gas to nitrometer.

The apparatus is cooled. Reservoir bulb of nitrometer is raised so that level of KOH becomes the same in reservoir bulb and nitrometer which means pressure becomes equal to atmospheric pressure. (MPBoardSolutions.com) Now, the volume of nitrogen in nitrometer is noted. Temperature of reaction and atmospheric pressure is noted from barometer. Aqueous tension at that temperature is noted from tables.

Observations and calculation:
Let,

1. Weight of organic substance = W gm
2. Volume of moist N = V ml
3. Temperature = t°C
4. Atmospheric pressure = P mm of Hg
5. Aqueous tension at t°C = p mm
6. Pressure of dry nitrogen = (P – p) mm

Calculation of volume of nitrogen at N.T.P.:

Question 9.
In an organic compound C = 40%, O = 53.34 % and H = 6.66%. If the vapour density is 30. Determine the molecular formula?
Or
In an organic compound A, C = 40% and H = 6.66%. The vapour density of A is 30. It turns blue litmus red and can react with ash. When its sodium salt is heated with soda lime, the first member of paraffin series obtained. What is A?
Solution:
C = 40%, H = 6.66%
O = 100 – [40 + 6.66] = 100 – 46.66 = 53.34%
In organic compound A:

Empirical formula of A = H₂O
Empirical formula mass = 12 + 2 + 16 = 30
Molecular mass = 2 × Vapour density
= 2 × 30 = 60
 = \(\frac{60}{30}\) = 2
∴ Molecular formula = (CH₂O)₂
= C₂H₄O₂ or CH₃COOH.
It is CH₃COOH

Question 10.
Explain the following with example:

1. Simple distillation
2. Chromatography
3. Crystallization.

Answer:
1. Simple distillation:
This method is employed for the purification of those liquids which boil without decomposition and are associated with non – volatile impurities. Liquids which have a difference of 30 – 40°C in their boiling points are purified by this method. On heating the mixture, vapours of pure substance are formed which condenses as they pass through the air or water condenser. (MPBoardSolutions.com) The pure liquid collects in the receiver while the non – volatile impurities are left behind in the flask. Some glass beads are also added to the distillation flask to avoid bumping.

2. Chromatography:
The process by which different components of a mixture are separated by distributing in stationary or mobile phases on the basis of difference in adsorption abilities on any adsorbent, is called chromatography.

Adsorption chromatography:
It is also known as column chromatography. It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

The mixture to be separated and purified is first dissolved in a suitable nonpolar organic solvent like petroleum ether, benzene, chloroform, alcohol, etc. This solution is allowed to flow down an absorption column.

For using this technique there should be proper adsorption column. Adsorbents like activated magnesia, alumina, calcium carbonate, gypsum, etc. is filled in a hard vertical tube (adsorbent column.).

3. Crystallization:
The method by which crystals of a substance can be made is called crystallization. Solids can be separated and purified by crystallization. e.g., nitre, alum, copper sulphate, etc. Some impure solids which differ in solubility in the same solvent, their separations can be achieved by fractional crystallization. (MPBoardSolutions.com) “If two or more components of a mixture which differ in their solubilities are dissolved in a solvent in which their solubilities slightly differ, they can be separated by fractional crystallization.”

Suppose, two solids A and B are dissolved in a solvent. If solubility of A is less as compared to B, then first saturated solution of the mixture is prepared and is allowed to cool. During crystallization first less soluble substance A will crystallize out. (MPBoardSolutions.com) It is separated by filtration. After this crystals of more soluble substance B will separate out. By this technique, crystals of both can be separated. The substances so separated are further crystallized many times to get pure substances.

Question 11.
Write the IUPAC name of following compounds:

(a) CH₃CH = C(CH₃)₂
(b) CH₂ = CH – C = C – CH₂
(c)
(d)
(e)

Answer:

MP Board Class 11 Chemistry Important Questions

 

MP Board Class 12th Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Questions

General Principles and Processes of Isolation of Elements Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
In the extraction of Fe from haematite, lime stone acts as :
(a) Reducing agent
(b) Slag
(c) Gange
(d) Flux.
Answer:
(d) Flux.

Question 2.
Cupellation is used in the metallurgy of:
(a) Cu
(b) Ag
(c) Al
(d) Fe.
Answer:
(b) Ag

Question 3.
Which is the essential part of photographic plates and Films :
(a) AgNO₃
(b) Ag₂S₂O₃
(c) AgBr
(d)Ag₂CO₃
Answer:
(c) AgBr

Question 4.
Malachite is :
(a) Cu₂S
(b) CUCO₃.CU(OH)₂
(c) Cu₂O
(d) CuCO₃.
Answer:
(b) CUCO₃.CU(OH)₂

Question 5.
AgBr is soluble in hypo because its forms :
(a) Ag₂SO₃
(b) Ag₂S₂O₃
(c) [Ag(S₂O₃)₃]^–
(d) [Ag(S₂O₃)₂]^3-
Answer:
(c) [Ag(S₂O₃)₃]^–

Question 6.
AgCl is soluble in ammonia due to the formation of:
(a) [Ag(NH₃)₄]⁺
(b) [Ag(NH₃)₃]²⁺
(c) [Ag(NH₃)₄]³⁺
(d) [Ag(NH₃)₂]⁺
Answer:
(d) [Ag(NH₃)₂]⁺

Question 7.
When KI mixed with solution of CuSO₄, it forms :
(a) Cul₂
(b) Cul^2-₂
(c) K₂[CuI₄]
(d) Cu_2& I₂ + I₂
Answer:
(d) Cu_2& I₂ + I₂

Question 8.
When KCN reacts with solution of CuSO₄, it forms :
(a) CU(CN)₂
(b) CuCN
(c) K₂[Cu(CN)₄]
(d) K₃[Cu(CN)₄]
Answer:
(d) K₃[Cu(CN)₄]

Question 9.
Na₂S₂O₃ used in photography as a :
(a) Reducing agent
(b) Developer
(c) Fixer
(d) Toning agent
Answer:
(c) Fixer

Question 10.
Calomel is :
(a) Hg₂ Cl₂
(b) HgCl₂
(c) Hg₂ Cl₂ +Hg
(d) Hg + Hgcl₂
Answer:
(a) Hg₂ Cl₂

Question 11.
German silver is an alloy of :
(a) Cu, Zn and Ni
(b) Cu, Zn and Sn
(c) Ag, Cu and Au
(d) Fe, Cr, Ni
Answer:
(a) Cu, Zn and Ni

Question 12.
Heating pyrites to remove sulphur is called :
(a) Roasting
(b) Liquation
(c) Calcination
(d) Smelting
Answer:
(a) Roasting

Question 13.
Reducing agent in the extraction of Iron from haematite is :
(a) CO
(b) C
(c) CO₂
(d) FeO
Answer:
(a) CO

Question 14.
The Chemical Composition of “Cryolite” mineral is :
(a) Al₂O₃
(b) Al₂O₃.12H₂O
(c) KAl Si₃O₈
(d) Na₃AlF₆
Answer:
(d) Na₃AlF₆

Question 15.
In Photography we use :
(a) Agl
(b) NH₃
(c) AgCl
(d) AgBr
Answer:
(d) AgBr

Question 16.
Blister copper is :
(a) Ore of copper
(b) Alloy of copper
(c) Pure copper
(d) Copper containing 1% empuritis.
Answer:
(d) Copper containing 1% empuritis.

Question 2.
Fill in the blanks :

1. Malachite is an ore of …………………
2. In stainless steel, along with iron ………………… and ………………… metals form alloys.
3.  ………………… is used as a purgative.
4. Colloidal solution of ………………… is used as a medicine of eyes.
5. AgNO₃ is known as …………………
6. Chemical formula of corrosive sublimate is …………………
7. Froath floatation process is generally employed for ………………… ores.
8. The process in which metal oxide is reduced by Al is known as …………………
9. Is used for drying ammonia …………………
10.  ………………… is called lunar caustic.
11. The chemical formula of fluorspar is …………………
12. Alkaline solution of HgCl₂ and Kl is known as …………………
13. Red hot steel is slowly cooled when it gets converted to soft steel, this is known as …………………

Answer:

1. Cu
2. Cr, Ni
3. Calomel
4. Ag
5. Lunar caustic,
6. HgCl₂
7. Sulphide
8. Aluminothermic
9. CuO
10. Silver Nitrate
11. CaF₂
12. Nesseler’s reagent
13. Annealing.

Question 3.
Match the following :
I

Answer:

1. (g)
2. (d)
3. (c)
4. (e)
5. (b)
6. (a)
7. (f)

II.

Answer:

1. (d)
2. (f)
3. (b)
4. (e)
5. (c)
6. (a)

Question 4.
Answer in one word / sentence :

1. The mixture of Cu₂S and FeS obtained from blast furnace is known as.
2. Which metal is purified by polling?
3. After developing which solution is used for the fixing of photographic films? Chemical formula of philosophers’s wool is.
4. Chemical formula of horn silver.
5. Which compound is normally used for toning in photography?
6. Which are known as coinage metals?
7. Give the name of ore used for extraction of Cu.
8. Give the name of ore used for extraction of Iron.
9. What is the name of graph which is drawn between the absolute temperature and standard free energy change for
10. formation of metal oxide?
11. What is the name of iron obtained after the removal of impurities from cast iron?
12. What is the method of slow cooling of hot hard steel known as?
13. What is the method of heating hard steel known as?
14. What is the method of heating steel in pressure of ammonia known as?
15. What is Lunar Caustic?

Answer:

1. Matte
2. Copper
3. Hypo solution (Na₂S₂O₃)
4. ZnO
5. AgCl
6. Aurric chloride
7. Cu, Ag and Au
8. Copperpyrite
9. Haematite
10. Ellinghum diagram
11. Wrought iron
12. Annealing
13. Softening
14. Nitriding
15. AgNO₃

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Important Questions

Haloalkanes and Haloarenes Very Short Answer Type Questions

Question 1.
What is plane polarized light?
Answer:
Light which vibrates in one specific plane is known as plane polarized light. On passing normal light through a Nicol prism, plane polarized light is obtained.

Question 2.
Write an example of 3° alkyl chloride.
Answer:
Tertiary butyl chloride

(2-chloro-2-methyl propane)

Question 3.
What is the order of boiling points of alkyl halides for the same alkyl group?
Answer:
Decreasing order of boiling points : RI > RBr > RCl > RF.

Question 4.
Write the equation of Swart’s reaction.
Answer:

Question 5.
What are Enantiomers?
Answer:
The stereoisomers related to each other as non – superimposable mirror images are called Enantiomers.

Question 6.
What are polyhalogen compounds? Give two examples.
Answer:
Carbon compounds which contain more than one halogen atom are known as polyhalogen compounds like CHCl₃, CCl₄ etc.

Haloalkanes and Haloarenes Short Answer Type Questions

Question 1.

1. Write Iodoform reaction.
2. Iodoform gives yellow ppt. with AgNO₃ solution but chloroform doesn’t Why?
3. What happens when ethyl bromide is heated with alcoholic KOH?

Answer:
1. Iodoform reaction:
When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH + 4I₂ + 6NaOH → 5Naf + HCOONa + 5H₂O + CHI₃

2. When iodoform is heated with AgNO₃ solution a yellow ppt. (Agl) is obtained but chloroform doesn’t give this reaction because in chloroform C – CI bond is more stable than C – I bondin iodoform.

3. On boiling Ethyl bromide with alcoholic KOH ethylene is formed.

Question 2.
Explain the Sandmeyer reaction with example.
Answer:
Decomposition of diazonium salts (Sandmeyer reaction):
When a diazonium salt solution is added to a solution of cuprous halide dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom.

Question 3.
Give chemical reaction between chlorobenzene and chloral in presence of cone. H₂SO₄.
Or, How is D.D.T. formed? Write its one application.
Answer:
DDT (Dichlorodiphenyl trichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone, H₂SO₄.

Application : It is an important pesticide.

Question 4.
What are Gem – dihalide and Vicinal – dihalide?
Answer:
When both halogen atoms are linked to one carbon atom of hydrocarbon then it is known as Gem-dihalide. Gem means geminal i. e., same position.

When both halogen atoms are connected to two different neighbouring carbon atoms, then it is known as vicinal dihalide. Vic means vicinal which means adjacent position.

Question 5.
What is Lucas reagent? Give its application.
Answer:
Solution of ZnCl₂ in cone. HCl is known as Lucas reagent.
Application:
It is used to differentiate primary, secondary and tertiary alcohols.
On adding the alcohol to Lucas reagent, a tertiary alcohol reacts immediately forming a ppt. of alkyl chloride. If the ppt. appears after few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 6.
Explain, Carbylamine reaction and give one application of this reaction. (MP 2018)
Answer:
Carbylamine reaction: On heating chloroform with primary amine (e.ganiline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

Application : Chloroform and primary amine can be tested by this reaction.

Question 7.
What is 666 (lindane)? Explain its preparation and use in agriculture.
Answer:
It is 1, 2, 3, 4, 5, 6 – HexachIorocyciohexane. It is obtained by heating benzene with chlorine in presence of sunlight.

Preparation : It is prepared by the chlnncriUiv benzene in the presence of ultraviolet light.

1,2, 3,4, 5, 6 – Hexachlorocyclohexane (B.H.C.)

Uses:
Benzene hexachloride is an addition compound and its γ – isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 8.
Explain the following reaction of chlorobenzene:

1. Ration with chlorine in the presence of FeCI3 in dark
2. Fittig reaction.

Answer:
1. When Chlorobenzene reacts with Cl₂ hi the presence of FeCl₃ in dark. o – dichlorobenzene and p – dichlorobenzene is obtained.

2. Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 9.
Write short notes on :

1. Hunsdiecker method and
2. Raschig process.

Answer:
1. Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.

2. Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂ (catalyst) at 230°C, chlorobenzene is formed.

Question 10.
Write method of preparation, properties and uses of Freon.
Answer:
Freon : Dichloro, Difluoro methane.
it is formed by the action of SbF₃ with CCl₃ in presence of SbCl₅.

It has very low boiling point due to which by increasing the pressure at room temperature it can be liquefied. It is a non – poisonous, non – combustible and inactive substance which is used as a cooling agent in the refrigerator. It is used in aerosol and foam.

Question 11.

1. B.P. of ethyl iodide is higher than b.p. of ethyl bromide. Give reason.
2. Explain why the m.p. of para dichlorobenzene is higher than its ortho arid meta derivatives.

Answer:

1. In alkyl halides containing same alkyl group boiling point increases with increase in atomic weights of halogen atoms. Molecular weight of ethyl iodide is more than ethyl bromide and therefore boiling point of ethyl iodide is also high.
2. Para derivatives of dichlorobenzene is more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 12.
Explain Friedel – Craft’s reaction with chemical equation,
Answer:
Friedel – Craft’s reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminium chloride to give alkyl benzene.

Acetylation:
Acetyl chloride reacts with benzene in presence of anhydrous aluminium chloride to give acetophenone.

Question 13.
Give the uses of freon – 12, D.D.T., Carbon tetrachloride and Iodoform. (NCERT)
Answer:
It is an important pesticide.
Freon – 12:
Since freons have been found to be one of the factors responsible for the depletion of ozone layer, they are being replaced by other harmless compounds in many countries.

D.D.T:
DDT is highly toxic and has strong insecticidal properties and thus it was widely used as an insecticide and pesticide.

Carbon Tetrachloride:
It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans. It is used as feed stock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing and general solvents use. Until the mid 1960’s, it was widely used as a cleaning fluid both in industry as a degreasing agent and in the home, as a spot remover and as fire extinguisher.

Iodoform:
It was earlier used as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.

Question 14.
Name of the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides : (NCERT)

1. (CH₃)₂CHCH(CI)CH₃
2. CH₃CH₂CH(CH₃)CH(C₂H)CI
3. CH₃CH₂C(CH₃)₂CH₂I
4. (CH₃)₃CCH₂CH(Br)C₆H₅
5. CH₃CH(CH₃)CH(Br)CH₃
6. CH₃C(C₂H₅)₂CH₂Br
7. CH₃C(CI)(C₂H₅)CH₂CH₃
8. CH₃CHC(Cl)CH₂CH(CH₃)₂
9. CH₃CH=CHC(Br)(CH₃)₂
10. p – ClC₆H₄CH₂CH(CH₃)₂
11. m – ClCH₂C₆H₄CH₂C(CH₃)₃
12. o – Br – C₆H₄CH(CH₃)CH₂CH₃.

Answer:

1. 2 – Chloro – 3 – methylbutane (2° alkyl)
2. 3 – Chloro – 4 – methyIhexane (2° alkyl)
3. 1 – Iodo – 2,2 – dimethylbutane (1° alkyl)
4. 1 – Bromo – 3,3 – dimethyl – 1 – phenylbutane (2° benzylic)
5. 2 – Bromo – 3 – methylbutane (2° alkyl)
6. 3 – Bromomethyl – 3 – methylpentane (1° alkyl)
7. 3 – Chloro – 3 – methylpentane (3° alkyl)
8. 3 – Chloro – 5 – methylhex – 2 – ene (vinyl)
9. 4 – Bromo- 4 – methylpent – 2 – ene (allylic)
10. 1 – Chloro – 4 – (2′-methylpropyl) benzene (aryl) or p – Chloro isobutyl benzene
11. 1 – Chloromethyl-3-(2’2′-diethylpropyl) benzene (benzylic) or m – Neopentyl benzyl chloride
12. 1 – Bromo – 2 – (l’ – methylpropyl) benzene (aryl).

Question 15.
Identify ‘A’, ‘B’, ‘C’ and ‘D’.

Answer:

Question 16.
Write the equation of following reactions of chlorobenzene : (MP 2015)

1. Halogenation
2. Nitration
3. Sulphonation
4. Alkylation.

Answer:
1. Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl₃ to form ortho and para substituted dihaloarene.

2. Nitration : Haloarenes react with nitrating mixture to form o – nitro and p – nitro substituted haloarenes.

3. Sulphonation : On heating with cone. H₂SO₄, 2 – Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

4. Alkylation : Alkylation takes place with alkyl halide in presence of anhydrous A1C13.

Haloalkanes and Haloarenes Long Answer Type Questions

Question 1.
Explain the nucleophilic substitution reaction in alkyl halide by S_N1 and S_N2 medianism.
Answer:
Nucleophilic substitution reaction:
In the carbon halogen bond ofhaloalkane. halogen atom is more electronepativ0 as compared to the carbon atom hence, the shared pair of electrons between carbon and halogen is more attracted by the halogen atom. As a result a small negative charge and an equivalent positive charge develops on halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial positive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.

The order of reactivity of different alkyl halide towards nucleophilic substitution reaction is:
RI > RBr > RC1 > RF

Mechanism of Nucleophilic substitution reactions :
Nucleophilic substitution reaction occurs through two different mechanism :

(1) S_N1 Mechanism (Unimolecular nucleophilic substitution) : In this mechanism following steps are involved :
(a) Formation of carbocation by dissociation of substrate i.e., reactant molecule.

(b) Attack of nucleophile on carbocation forming the product.

(2) S_N2 mechanism (Bimolecular nucleophilic substitution):
Reactions of this type occur in one step i.e. they are concerted reactions. These reaction nucleophilic attack results in a transition state in which both the reactant molecules are partially bonded to each other and then the halide ion escapes out forming the product.
ROH^– + CH₃X → CH₃OH + RX

Rate of reaction = K[(RX)OH^–]
Order of reactivity of alkyl halide is : Primary > Secondaxy > Tertiary.

Question 2.
Draw labelled diagram of laboratory method for preparation of iodoform from akfohol. Write related chemical equation.
Answer:

Chemical reaction:

Question 3.
Write the structure of the major organic product in each of the following reactions: (NCERT)

Answer:

Question 4.
Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labelled diagram, equation and principle.
Answer:
Laboratory method : Chloroform is prepared in the laboratory by the action of water and bleaching powder on ethyl alcohol or acetone.

Method:
About 100 gm of bleaching powder made into a paste by adding about 200 ml of water and taken in a flask fitted with a condenser. Now, 25 ml of alcohol or acetone is added and the mixture is distilled, chloroform collects as a heavy liquid under water.

It is washed with dilute NaOH solution then with water, dried over fused calcium chloride and redistilled. The available chlorine of bleaching powder acts as oxidising as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.
CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

The chemistry involved in the conversion of alcohol and acetone into chloroform is as shown below:

(A) From alcohol: The steps involved are :
(i) Ethyl alcohol is oxidized by chlorine to acetaldehyde.

(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloro acetaldehyde.

(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.

Calcium hydroxide Chloral Chloroform Calcium formate.

Question 5.
MaloaIkanes are more reactive than haloarenes. Give reason.
or,
Why, aryl halides are less reactive than alkyl halides?
Answer:
In aryl halides, halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slowly than alkyl halides. There is two reasons for the less reactivity of aryl halides.
(i) In aryl halides sp³hybridization takes place whereas in alkyl halides sp² hybridization is present due to sp² hybridization in haloarenes the halogen are attached to nucleus more strongly.

(ii) Due to the presence of resonance in aryl halides there is some double bond character in C – Cl bond. Thus, the bond length of C – Cl bond is lesser than C – Cl bond in haloalkanes. Therefore, it is difficult to replace the halogen of haloarenes.

Question 6.
What product is formed by the reduction of chloroform? Give the chemical equation when it reacts to nitric acid and acetone.
Answer:
Reduction:
1. On heating with Zn and HCl, it reduces to form methylene dichloride.

2. On heating with zinc dust and water, methane is formed.

Reactions of Chloroform :
1. With Conc.HNO₃:
On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

2. With Acetone:
Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.

Question 7.
Give the Chemical reaction when chloroform reacts with following:

1. Oxidation
2. Carbylamine reaction
3. Ag powder
4. Nitration
5. Reimer – Tiemann reaction.

Or
How will you obtain the following from chloroform:

1. Carbonyl chloride
2. Acetylene
3. Chloropicrin
4. Phenyl isocyanide
5. Chloretone
6. Salicylaldehyde.

How trichloro methane reacts with :

1. Atmospheric air
2. Aniline and ale. KOH
3. Ag powder
4. Cone. HNO₃
5. Phenol
6. Acetone.

Answer:
(a) Action of air and light (Oxidation):
Chloroform oxidizes in presence of sunlight and air and forms a poisonous gas, phosgene (carbonyl chloride).

Ordinary chloroform contains phosgene gas and is used as a solvent. Pure chloroform which is used as an anaesthetic does not contain even traces of phosgene. While preserving chloroform which is to be used as an anaesthetic, the following precautions are taken:
(i) The chloroform is filled in blue or brown coloured bottle up to the neck. After putting a stopper, the bottle is kept in dark. As there is no empty space in the bottle, it is also to from air

(ii) One percent ethyl alcohol is added in the bottle. If phosgene gas is formed, alcohol reacts with it to form diethyl carbonate, a non – toxic substance, i.e. (C₂H₅)₂CO₃

(b) Carbylamine reaction:
On heating chloroform with primary amine (example aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

(c) Reaction with Ag powder or Dehalogenation: On heating chloroform with silver powder, pure acetylene gas is formed.
CHC₁₃ + 6Ag + CI3CH → HC ≡ CH + 6 AgCl

(d) Nitration:
Haloarenes react with nitrating mixture to form o – nitro and p – nitro substituted haloarenes.

(e) Reimer – Tiemann reaction:
On heating chloroform with concentrated alkali and phenol at 60 – 70°C, o – hydroxy benzaldehyde (salicylaldehyde) is formed. Traces of p- hydroxybenzaldehyde are also formed.

(f) Acetone :
Chloroform condenses with acetone in presence of NaOH to form chloretone which is a hypnotic of high grade.

Question 8.
An alcohol ‘A’ on reaction with cone. H₂SO₄ gives an alkene ‘B’. ’B’ after bromination with sodamide gives dehydrogenated compound ‘C’. ‘C’ on reaction of H₂SO₄ in presence of H₂SO₄ gives ‘D’. Identify ‘A’, ‘B% ‘C’, and ‘D’. (MP 2017)
Answer:

MP Board Class 12th Chemistry Important Questions